CBSE Class 12 Physics Capacitance Theory and Examples

CBSE Class 12 Physics Capacitance Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

CAPACITANCE - THEORY AND SOLVED EXAMPLES

Class 12 Physics notes and solved examples on capacitance, covering capacitance of an isolated sphere, energy of a charged conductor, distribution of charge between conductors, the parallel plate, spherical and cylindrical capacitors, dielectrics, combinations of capacitors, energy stored in a capacitor, charging and discharging of a capacitor with time constant RC, and worked examples based on each topic.

1. CAPACITANCE

(i) When a conductor is given a charge, its potential rises in proportion to the charge given, i.e. Q ∝ V, or Q = CV, or C = Q / V. Here C is a constant and is called the CAPACITANCE of the conductor.
(ii) Electrical capacitance of a conductor is a measure of its ability to hold electric charge.
(iii) A given conductor can be charged only up to a limit. Charging beyond the limit ionises the medium, and the charge leaks into the medium.
(iv) C depends on (a) the size and shape of the conductor, (b) the surrounding medium, and (c) the presence of other conductors nearby.
(v) Units: (1) coulomb / volt, (2) farad. 1 farad = 1 coulomb / volt.
(vi) If the potential of a conductor rises by 1 volt when it is charged by 1 coulomb, its capacitance is 1 farad.
(vii) The farad is too large a unit: a sphere of 1 farad must have a radius of 9 × 10⁹ m. In practice, these units are used: 1 μF = 1 × 10^-6 F, 1 nF = 1 × 10^-9 F, 1 pF = 1 × 10^-12 F.
(viii) Dimension of capacitance: [M^-1 L^-2 T⁴ A²].
(ix) CGS unit: 9 × 10¹¹ statfarad = 1 farad.

Example based on Capacitance

Ex.1: A conductor gets a charge of 60 pC when it is connected to a battery of emf 6 mV. Then the capacitance of the conductor is...........
(A) 120 × 10^-6 (B) 1.2 × 10^-9
(C) 10 × 10^-9 F (D) 10 × 10^-6 F

Answer: (C) 10 × 10^-9 F

1. The definition of capacitance is C = Q / V.
2. C = (60 × 10^-12) / (6 × 10^-3) = 10 × 10^-9 F = 10^-8 F.
3. Hence, the correct option is (C).

Teacher's Note:
a) Convert pC and mV to base units before dividing.
b) 10^-8 F is 10 nF, which is the same as option (C).

Ex.2: Farad is not equivalent to- (A) Q / V (B) J / V² (C) Q² / J (D) CV²

Answer: (D) CV²

1. Capacitance C = Q / V, so the farad equals C = Q / V (option A).
2. Energy U = ½CV² = Q² / 2C = ½QV, so C = Q² / J and C = J / V² (options B and C) are equivalent to the farad.
3. CV² has the unit of joule, not farad, so it is not equivalent to the farad.
4. Hence, the answer is (D).

Teacher's Note:
a) Check each option by putting it in terms of coulomb and volt.
b) CV² has the dimensions of energy.

1.1 Capacitance of an isolated spherical conductor

Q = charge on the sphere, V = potential at the surface of the sphere, R = radius of the sphere.
(i) V = Q / (4πε₀R), so C = Q / V = 4πε₀R.
(ii) C = 4πε₀R, hence Q / V = C = 4πε₀R (in MKS) = R (in CGS).
(iii) If the permittivity of the medium is K, then C_m = 4πε₀KR.
(iv) C ∝ R and C ∝ K.
(v) C does not depend on the charge given to the conductor.
(vi) Units of ε₀ = farad / metre, so ε₀ = C / (4πR).
Note: C² / (N m²) is another unit of ε₀. Thus F/m and C² / (N m²) both represent the same physical quantity.

Ex.3: Can a sphere of radius 1 cm, placed in air be given a charge of 1 coulomb. Potential of sphere of 1 cm radius is.

Answer: No

1. Potential of the sphere: V = kq / r = 9 × 10⁹ × 1 / (10^-2) = 9 × 10¹¹ V.
2. This potential is so high that the surrounding air gets ionised, since air breaks down at about 3 × 10⁶ V per metre.
3. The charge leaks into the air, so the sphere cannot hold 1 coulomb.

Teacher's Note:
a) Ionisation of air limits the charge a conductor can hold.
b) A small sphere can hold only a tiny charge before leakage starts.

Ex.4: Two spheres, one being hollow and another solid have equal radius. If they are charged to V volt, then which one will acquire more charge?

Answer: Both will acquire the same charge.

1. Capacitance of a sphere depends only on its radius, C = 4πε₀R.
2. Both spheres have equal radius, so their capacitances are equal.
3. Charge Q = CV, so both acquire the same charge at the same potential.

Teacher's Note:
a) The inside of a conductor does not affect its capacitance.
b) Hollow and solid spheres of the same radius have the same capacitance.

Ex.5: If the diameter of earth is 128 × 10² km, then its capacity will be- (A) 111 μF (B) 211 μF (C) 331 μF (D) 711 μF

Answer: (D) 711 μF

1. Radius of earth R = diameter / 2 = 64 × 10² km = 6.4 × 10⁶ m.
2. C = R / k = 6.4 × 10⁶ / (9 × 10⁹) = 7.11 × 10^-4 F = 711 μF, where k = 1 / 4πε₀.
3. Hence, the answer is (D).

Teacher's Note:
a) Use the radius, not the diameter, in C = 4πε₀R.
b) Convert km to metres before dividing.

Ex.6: 64 water drops having equal charges & equal radius combine to form one bigger drop. The capacity of bigger drop, as compared to that of smaller drop will be- (A) 8 times (B) 64 times (C) 4 times (D) 16 times

Answer: (C) 4 times

1. Volume is conserved: (4/3)πR³ = 64 × (4/3)πr³, so R = 4r.
2. C = 4πε₀R, so C_bigger / C_smaller = R / r = 4.
3. Hence, the bigger drop has 4 times the capacity.

Teacher's Note:
a) Capacitance is proportional to radius, not volume.
b) Cube root of the number of drops gives the radius ratio: 64^1/3 = 4.

Ex.7: Capacity of a metal sphere will be 1 μf, if it's radius is nearly- (A) 9km (B) 64 km (C) 81 km (D) None of these

Answer: (A) 9 km

1. C = 4πε₀R, so R = C / (4πε₀) = 9 × 10⁹ × 1 × 10^-6 = 9 × 10³ m = 9 km.
2. Hence, the answer is (A).

Teacher's Note:
a) Multiply by 9 × 10⁹ to find R from C in farads.
b) A capacitance of 1 μF needs a sphere as large as a small town.

Ex.8: A radioactive source in the form of a metal sphere of diameter 10^-3 m emits β paraticles at a constant rate of 6.25 × 10¹⁰ particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that 80% of emitted β Particles escape from the surface.

Answer:
1. Radius r = 0.5 × 10^-3 m, so C = 4πε₀r = (0.5 × 10^-3) / (9 × 10⁹) = (1/18) × 10^-12 F.
2. Rate of escape of charge = (80 / 100) × 6.25 × 10¹⁰ × 1.6 × 10^-19 = 8 × 10^-9 C/s.
3. Charge needed for a rise of 1.0 V: q = CV = (1/18) × 10^-12 × 1.0 C.
4. Time t = q / (8 × 10^-9) = 10^-12 / (18 × 8 × 10^-9) = 6.95 × 10^-6 s.
5. Hence, t = 6.95 μs.

Teacher's Note:
a) Only the escaping 80% of the charge leaves the source, so use 0.8 times the emission rate.
b) Use the radius, which is half the diameter.

2. ENERGY OF A CHARGED CONDUCTOR

(i) The work done in charging a conductor is stored as potential energy in the electric field in the vicinity of the conductor.
(ii) Suppose a conductor is given a charge q, so its potential is V = q / C. Work done to bring a further charge dq is dW = V dq = (q / C) dq. So the work done to charge it from 0 to Q is W = ∫₀^Q (q / C) dq = Q² / 2C = ½CV² = ½QV. This energy is U, which is stored.
(iii) This energy does not depend on the size of the conductor.

Ex.9: Two charged conductor having capacities 2C & C, potentials V & V/2 respectively. The ratio in their energies will be- (A) 1 : 2 (B) 2 : 1 (C) 8 : 1 (D) 1 : 8

Answer: (C) 8 : 1

1. E₁ = ½(2C)V² = CV².
2. E₂ = ½C(V / 2)² = CV² / 8.
3. E₁ : E₂ = CV² : CV²/8 = 8 : 1.
4. Hence, the answer is (C).

Teacher's Note:
a) Energy depends on the square of the potential.
b) Halving the potential reduces energy to one-fourth, and doubling the capacity doubles it.

3. DISTRIBUTION OF CHARGES

(i) Two insulated conductors A and B with capacitances C₁ and C₂ are given charges q₁ and q₂ and raised to potentials V₁ and V₂ respectively. Then q₁ = C₁V₁ and q₂ = C₂V₂.
(ii) When these conductors are joined by a thin wire, positive charge flows from the conductor at higher potential to the conductor at lower potential till their potentials become equal.
(iii) Charge remains conserved in this process, i.e. if q'₁ and q'₂ are the charges after distribution and V the common potential, then q'₁ = C₁V, q'₂ = C₂V, and C₁V₁ + C₂V₂ = C₁V + C₂V.
(iv) V = (total charge) / (total capacitance) = (C₁V₁ + C₂V₂) / (C₁ + C₂).
(v) On connecting two charged conductors, their distributed charges are in the ratio of their capacitances: q₁ / q₂ = C₁ / C₂.
(vi) Loss of energy: ΔU = U_initial - U_final = ½ (C₁C₂ / (C₁ + C₂)) (V₁ - V₂)².
(vii) Like conductors, there is redistribution of charge and consequently loss of energy on connecting two condensers.
Note: (Very Important): (i) When positive plate of a condenser is connected to positive plate of a second condenser, then Q₁, Q₂, V₁, V₂ are taken to be positive. (ii) When positive plate of a condenser is connected to negative plate of the second condenser, then values of Q₁, Q₂, V₁, V₂ are taken with appropriate sign.
Special case: If the second conductor was initially uncharged, i.e. q₂ = 0, V₂ = 0, then C₁ is connected in parallel to conductor of capacity C₂. The common potential (after equilibrium) is V = C₁V₁ / (C₁ + C₂). The loss of energy becomes ΔU = (1/2) (C₁C₂) / (C₁ + C₂) V₁². The percentage of energy loss in this case = (ΔU / U_i) × 100.
This energy appears partly as heat in the connecting wire and partly as light and sound if sparking occurs.
Note: If V₁ = V₂, then ΔU = 0, i.e. there is no loss of energy on connecting two conductors at the same potential.

Example based on Distribution of charges

Ex.10: The radii of two charged metallic spheres are 15 cm and 10cm. Each has a positive charge of 150 micro coulomb. They connected by a conducting wire. Calculate the common potential of the spheres after connecting and the quantity of charge on each.

Answer:
1. Capacity of the first sphere: C₁ = (1 / 9.0 × 10⁹) × 0.15 = (15/9) × 10^-11 F.
2. Capacity of the second sphere: C₂ = (1 / 9.0 × 10⁹) × 0.10 = (10/9) × 10^-11 F.
3. Charge on each sphere: q₁ = q₂ = 150 × 10^-6 C, so total charge = 300 × 10^-6 C.
4. Common potential: V = (q₁ + q₂) / (C₁ + C₂) = (300 × 10^-6) / ((25/9) × 10^-11) = 1.08 × 10⁷ V.
5. Charge on the first sphere: q₁ = C₁V = (15/9) × 10^-11 × 1.08 × 10⁷ = 180 × 10^-6 C.
6. Charge on the second sphere: q₂ = C₂V = (10/9) × 10^-11 × 1.08 × 10⁷ = 120 × 10^-6 C.

Teacher's Note:
a) Charges on the spheres redistribute in the ratio of their capacitances.
b) The total charge stays the same: 180 + 120 = 300 μC.
c) The common potential depends on the total charge and the total capacitance.

Ex.11: Two conductors having capacities 2μF & 5μF and potentials 2 volt and 10 volt respectively. The ratio in their charges after connecting by a wire will be- (A) 2 : 5 (B) 5 : 2 (C) 1 : 5 (D) 5 : 1

Answer: (A) 2 : 5

1. After connecting, charges are distributed in the ratio of capacitances: q₁ / q₂ = C₁ / C₂.
2. q₁ / q₂ = (2 × 10^-6) / (5 × 10^-6) = 2 / 5.
3. Hence, the answer is (A).

Teacher's Note:
a) The ratio of charges after connection depends only on the capacitances.
b) The potentials of the conductors do not affect the final ratio.

Ex.12: Two isolated metallic solid spheres of radii R and 2R are charged such both of these have same charge density σ. The spheres are located far away from each other and connected by a thin conduction wire. Find the new charge density on the bigger sphere.

Answer:
1. Let the charges on the spheres of radii R and 2R be Q₁ and Q₂. Equal surface densities give Q₁ / 4πR² = Q₂ / 4π(2R)², so Q₂ = 4Q₁.
2. Total charge Q = Q₁ + Q₂ = 5Q₁.
3. After connection, charges are in the ratio of capacitances (R : 2R), so Q'₁ = 5Q₁ / 3 and Q'₂ = 10Q₁ / 3.
4. New charge density on the bigger sphere: σ' = Q'₂ / (4π(2R)²) = (10Q₁ / 3) / (16πR²).
5. Substituting Q₁ = 4πR²σ gives σ' = (5 / 6)σ.

Teacher's Note:
a) Equal surface density means the charge is proportional to the area.
b) Charge is conserved on connection, so the total charge 5Q₁ must be used.
c) Surface density falls on the bigger sphere because its radius doubles.

4. CAPACITOR / CONDENSER

As we know, that capacitance of a conductor depends on the presence of other conductors nearby. This fact is used to make a capacitor.
(i) Concept of condenser: two conductors are brought close together, so that they have equal and opposite charges. These conductors are called the PLATES of the capacitor. Capacitance is C = Q / V_A.
(ii) Conductors in a condenser are arranged in such a way that capacity of conductors is increased without bringing any change in their size.
(iii) For a condenser, Q = CV, where Q is the charge given to any one conductor and V is the potential difference between the conductors.
(iv) In fact, a capacitor is a pair of two conductors of any shape which are close to each other and have equal and opposite charges.
(v) Capacitance of a condenser depends on the following points.
(a) Distance between plates: C is inversely proportional to d, i.e. C ∝ 1/d. As d increases, C decreases and vice versa.
(b) Overlapping area of plates: C is directly proportional to A, i.e. C ∝ A.
(c) Medium between the plates: capacitance is directly proportional to the dielectric constant K of the medium, i.e. C ∝ K.
(vi) If capacitance of a condenser in air or vacuum is C, then when its medium (K) is a dielectric, the capacitance becomes KC.
(vii) Condensers provide infinite resistance for a D.C. while they allow A.C. to pass through them (provide limited/finite resistance).
Note: A parallel plate capacitor is charged and the battery is then removed. When the distance between the plates is reduced: the charge remains the same, the capacitance increases, the potential difference decreases, the electric field does not change, and the energy decreases because work is done in bringing the plates closer. When a dielectric is placed between the plates: the charge remains the same, the capacitance increases, the potential difference decreases, the electric field decreases, and the energy is reduced.

Now a second conductor B is brought near A:

V'_A = V_A + V_B− + V_B+. V_B− is negative, V_B+ is positive and |V_B−| > |V_B+|, because of the lesser distance of the negative charge. ⇒ V'_A < V_A ⇒ capacitance increases.

If B is earthed, V_B+ = 0, hence V'_A = V_A + V_B− and the capacitance of A increases.

Ex.13: A gold leaf electroscope is positively charged and an earthed conducting plate is brought near it, then- (A) capacity remains constant, potential decreases, charge increases (B) capacity decreases, potential decreases, charge remains constant (C) capacity increases, potential decreases, charge remains constant (D) capacity remains constant, potential decreases, charge increases

Answer: (C) capacity increases, potential decreases, charge remains constant

1. The earthed plate brought near the charged electroscope increases its capacitance.
2. Charge on the electroscope stays the same, since it is not connected to any source.
3. Since C = Q / V, the potential V decreases when C increases.
4. Hence, the answer is (C).

Teacher's Note:
a) An earthed plate nearby increases capacity by bringing the charges closer.
b) The charge stays constant because the electroscope is isolated, so the capacity rise lowers the potential.

Ex.14: The net charge on a condenser is- (A) 2q (B) q/2 (C) zero (D) ∞

Answer: (C) zero

1. The two plates of a capacitor carry equal and opposite charges, +q and -q.
2. Net charge = +q + (-q) = 0.
3. Hence, the answer is (C).

Teacher's Note:
a) The capacitor stores energy in the field, not as net charge.
b) Each plate carries charge of magnitude q, with opposite signs.

Ex.15: The capacitance of a capacitor is C when the distance between the plates is halved, the capacitance will becomes- (A) half (B) twice (C) one-fourth (D) four-times

Answer: (B) twice

1. C ∝ 1/d, so C' / C = d / (d/2) = 2.
2. Hence, the capacitance becomes twice.

Teacher's Note:
a) Halving the separation doubles the capacitance.
b) Check the direction: smaller separation means larger capacitance.

Ex.16: If the distance between the plates of a capacitor is made half and the area of plates is doubled, then the capacitance will becomes- (A) Twice (B) Four times (C) Half (D) One-fourth

Answer: (B) Four times

1. C ∝ A / d.
2. C' / C = (2A / (d/2)) / (A / d) = 2 × 2 = 4.
3. Hence, the capacitance becomes four times.

Teacher's Note:
a) Both changes increase the capacitance in the same direction.
b) Write the ratio of the areas and distances separately, then multiply.

Ex.17: If 50μF be the capacity of a capacitor in air, and 110μF in of oil then the dielectric constant oil will be- (A) 0.45 (B) 0.55 (C) 1.10 (D) 2.20

Answer: (D) 2.20

1. C ∝ K, so C_oil / C_air = K_oil / K_air.
2. K_oil = (110 × 10^-6) / (50 × 10^-6) = 2.2.
3. Hence, the answer is (D).

Teacher's Note:
a) Air has K = 1, so the oil constant equals the ratio of capacitances.
b) A dielectric constant is always greater than 1.

5. ENERGY STORED IN A CONDENSER

(i) If C = capacitance of condenser, Q = charge on condenser, V = potential it is raised to, then E = ½CV² = Q² / 2C = ½QV. This energy is stored as the electric field between plates. The energy density of the field is ½ε₀E².
(ii) Energy density of a field is u = ½ε₀K²E² in the medium, where u is called the energy density of the medium.
(iii) The energy does not depend on the size of the conductor.

Ex.18: Two capacitor C₁ and C₂ have equal amount of energy stored in them. The ratio of p.d. across their plates is- (A) √(C₂/C₁) (B) √(C₁/C₂) (C) 1 : 2 (D) 2 : 1

Answer: (A) √(C₂/C₁)

1. Equal energies: ½C₁V₁² = ½C₂V₂².
2. So V₁² / V₂² = C₂ / C₁, and V₁ / V₂ = √(C₂ / C₁).
3. Hence, the answer is (A).

Teacher's Note:
a) The smaller capacitor must have the larger voltage for equal energy.
b) Take the square root only at the end.

Ex.19: If the distance between the plates of a capacitor is d and potential difference is V then the energy density will be- (A) ½ε₀V²/d² (B) ½ε₀V/d (C) ½ε₀V²d² (D) ½ε₀V·d

Answer: (A) ½ε₀V²/d²

1. Energy density = ½ε₀E².
2. E = V / d, so energy density = ½ε₀(V / d)² = ½ε₀V² / d².
3. Hence, the answer is (A).

Teacher's Note:
a) Energy density is energy per unit volume, so it depends on E² only.
b) Do not confuse energy density with total energy.

Ex.20: In the given fig, when the switch is open, the space between the capacitor is filled with dielectrics of constant 3. The ratio of energies capacitor before an after filling the dielectric will be- (A) 2/5 (B) 5/2 (C) 3/5 (D) 5/3

[Figure: Circuit with a battery V, a switch S, capacitor A (C) connected in parallel with capacitor B, and capacitor C connected in parallel with capacitor B. The space in capacitor A is filled with a dielectric of constant 3 when the switch is open.]

Answer: (C) 3/5

1. Initially, both capacitors have the same charge, q_A = q_B = CV.
2. When the dielectric is filled, C₁ = K C = 3C.
3. When the switch is opened, the charge on A stays the same, so q_B = CV = C₁V₁, giving V₁ = V / 3.
4. Initial energy: E_i = ½CV² + ½CV² = CV².
5. Final energy of A: ½(3C)V² = (3/2)CV².
6. Final energy of B: ½(3C)(V / 3)² = CV² / 6.
7. Total final energy E_f = (3/2)CV² + CV²/6 = (5/3)CV².
8. Ratio E_i / E_f = CV² / ((5/3)CV²) = 3/5.
9. Hence, the answer is (C).

Teacher's Note:
a) The printed working uses the same final charge on A while the dielectric is inserted with the switch open.
b) Energy of the system falls because the dielectric draws the field in.

Ex.21: Calculate the heat generated when a condenser of 100μF capacity and charged to 200 volts is discharged through a 2 ohm resistance. (A) 0 (B) 2 J (C) 1 J (D) 4 J

Answer: (B) 2 J

1. Energy stored: U = ½CV² = ½ × 100 × 10^-6 × (200)² = 2 J.
2. When discharged through a resistor, the whole energy is dissipated as heat.
3. Hence, heat produced Q = U = 2 J.

Teacher's Note:
a) The resistance value does not change the total heat, only the rate at which it is produced.
b) Convert μF to farads before using ½CV².

6. COMBINATION OF CAPACITORS

There are two possible combinations: (A) Series and (B) Parallel.

6.1 Series combination

(a) Charge on each capacitor is the same, i.e. Q = C₁V₁ = C₂V₂ = C₃V₃ = ...........
(b) Potential difference across each capacitor is inversely proportional to its capacity, i.e. V ∝ 1/C. So V₁ = Q / C₁, V₂ = Q / C₂, ...
(c) Total potential difference in the circuit is the sum of the potential differences across each capacitor, i.e. V = V₁ + V₂ + V₃ + ........... or V / Q = 1/C = 1/C₁ + 1/C₂ + 1/C₃ + ............
Note:
(i) If n equal capacitors are connected in series, their equivalent capacitance will be C/n.
(ii) If C₁, C₂, C₃ ............. all capacitors are connected in series then total energy U = ½q²[1/C₁ + 1/C₂ + 1/C₃ + ............].
Note:
1. Please keep in mind that you are asked to find C and not 1/C.
2. Equivalent capacitance is harmonic sum of the individual capacitance and this is lower than the lowest capacitance in the circuit.
(iii) Series combination is useful when: (1) a capacitance is needed which is lower than the lowest of given capacitors, or (2) the capacitance of the combination is to be increased at low voltage, where high capacitance is needed.
(vi) Series combination of two capacitors: C = C₁C₂ / (C₁ + C₂); V₁ = C₂V / (C₁ + C₂); V₂ = C₁V / (C₁ + C₂); Q = (C₁C₂ / (C₁ + C₂)) V = C₁V₁ = C₂V₂.

6.2 Parallel combination

(a) Potential difference across each capacitor is the same and is equal to the potential difference applied across the circuit.
(b) Charge on each capacitor is proportional to its capacity, i.e. Q₁ = C₁V, Q₂ = C₂V, Q₃ = C₃V, ........., so Q ∝ C.
(c) Total charge in the circuit Q = Q₁ + Q₂ + Q₃ + ......
(d) If C is the total capacitance of the circuit then C = C₁ + C₂ + C₃ + ............
(e) This combination is useful when the capacitance of the combination is to be increased. At low voltage high capacitance is needed.
(f) For parallel combination of two capacitors: C = C₁ + C₂; Q₁ = C₁Q / (C₁ + C₂); Q₂ = C₂Q / (C₁ + C₂); Q = CV.
(g) The total energy stored in parallel combination of two capacitors is U = U₁ + U₂ = ½C₁V² + ½C₂V² = ½V²(C₁ + C₂).
Note:
(i) If n capacitors of equal capacity C are connected in parallel, their equivalent capacity will be nC.
(ii) If n capacitors, each having capacitance C, are first connected in series and then in parallel, the ratio of resultant capacitances in the two cases will be C / nC = 1 / n².

Example based on Combination of capacitors

Ex.22: Three capacitors C₁ = 1 μF, C₂ = 2μF and C₃ = 3μF should be connected in following manner, so that the resultant capacitance becomes the least & (A) All in parallel (B) All in series (C) C₁ & C₂ in series C₃ in parallel (D) C₂ & C₃ in series & C₁ in parallel

Answer: (B) All in series

1. All in series: 1/C = 1/1 + 1/2 + 1/3 = 11/6, so C = 6/11 μF ≈ 0.55 μF.
2. All in parallel: C = 1 + 2 + 3 = 6 μF.
3. C₁ and C₂ in series give 2/3 μF, in parallel with C₃: 2/3 + 3 = 11/3 μF.
4. C₂ and C₃ in series give 6/5 μF, in parallel with C₁: 6/5 + 1 = 11/5 μF.
5. The least resultant is 6/11 μF, so the answer is (B).

Teacher's Note:
a) Series combinations always give a smaller capacitance than the smallest one.
b) Compare all four results before choosing.

Ex.23: Four capacitors 102 μF, 104 μF, 116 μF & 108 μF are connected in series, the resultant capacitance will be- (A) more than 116 μF (B) more than 102 μF but less than 116 μF (C) less than 102 μF (D) The average of all the four

Answer: (C) less than 102 μF

1. In series, the resultant capacitance is less than the least capacitance of the combination.
2. The least capacitance here is 102 μF, so the resultant is less than 102 μF.
3. Hence, the answer is (C).

Teacher's Note:
a) The resultant of a series combination is always below the smallest capacitor.
b) An average of the capacitances has no role in series combinations.

Ex.24: In the following fig. three capacitors (each of 9μF) are connected, the equivalence capacity will be- (A) 3 μF (B) 27 μF (C) 1 μF (D) 3 F

[Figure: Three capacitors C₁, C₂ and C₃ connected in a single row between two terminals, with the voltage V applied across the whole row.]

Answer: (A) 3 μF

1. The three 9 μF capacitors are in series: 1/C = 1/9 + 1/9 + 1/9 = 1/3.
2. Hence, C = 3 μF.

Teacher's Note:
a) Three equal capacitors in series give one-third of one capacitor.
b) The unit remains μF.

Ex.25: In the following problem, the correct order of p.d. across the plates of capacitors will be- (A) V₁ = 360 volt, V₂ = 40 volt, V₃ = 90 volt, (B) V₁ = 360 volt, V₂ = 40 volt, V₃ = 90 volt, (C) V₁ = V₂ = V₃ = 40 volt, (D) V₁ = 90 volt, V₂ = 360 volt, V₃ = 40 volt

In the PDF the four options are: (A) V₁ = 360 volt, V₂ = 40 volt, V₃ = 90 volt; (B) V₁ = 360 volt, V₂ = 360 volt, V₃ = 90 volt; (C) V₁ = V₂ = V₃ = 40 volt; (D) V₁ = 90 volt, V₂ = 360 volt, V₃ = 40 volt. (Option (B) printed above repeats (A) by mistake and should read as given here.)

Answer: (C) V₁ = V₂ = V₃ = 40 volt

1. The three capacitors are in series, so the charge on each is the same.
2. Equivalent capacitance: C = 3 μF, applied voltage 120 V.
3. Charge q = CV = 3 × 10^-6 × 120 = 360 × 10^-6 C.
4. V₁ = q / C₁ = 360 × 10^-6 / (9 × 10^-6) = 40 V, and likewise V₂ = V₃ = 40 V.
5. Hence, the answer is (C).

Teacher's Note:
a) In series, equal capacitors share the voltage equally.
b) Option (B) in the PDF is V₁ = 360 volt, V₂ = 360 volt, V₃ = 90 volt; only option (C) gives equal voltages.

Ex.26: An infinite no. of capacitors of capacitance C, 4C, 16C, 64C ∞ are connected in series, their resultant capacitance will be- (A) ∞ (B) C/2 (C) 4C/3 (D) 3C/4

Answer: (D) 3C/4

1. 1/C' = 1/C + 1/4C + 1/16C + ... = (1/C)(1 + 1/4 + 1/16 + ...).
2. Sum of the infinite geometric series with ratio 1/4 is 1 / (1 - 1/4) = 4/3.
3. So 1/C' = (1/C)(4/3), giving C' = 3C/4.
4. Hence, the answer is (D).

Teacher's Note:
a) Use the sum a / (1 - r) when the ratio r is less than 1.
b) Series combination of infinite capacitors gives a finite value.

Ex.27: If capacitors C, 3C, 5C .... ∞ in one network and 2C, 4C, 6C .... ∞ are connected in another network. If the effective capacitance of first network be C₁ and that of network second be C₂ then C₁C₂ / (C₂ - C₁) equal to- (A) C / log 2 (B) log 2 / C (C) ∞ (D) (log 2)² / C

Answer: (A) C / log_e 2

1. For the first network, 1/C₁ = 1/C + 1/3C + 1/5C + ........
2. For the second network, 1/C₂ = 1/2C + 1/4C + 1/6C + ........
3. Subtracting, 1/C₁ - 1/C₂ = (1/C)[1 - 1/2 + 1/3 - 1/4 + ........].
4. Using log_e(1 + x) = x - x²/2 + x³/3 - ........, the bracket equals log_e 2.
5. So (C₂ - C₁) / (C₁C₂) = log_e 2 / C.
6. Hence, C₁C₂ / (C₂ - C₁) = C / log_e 2, which is option (A).

Teacher's Note:
a) Reciprocals of series capacitances are added, which gives the difference in step 3.
b) Recognise the logarithm series, which is the key shortcut.

Ex.28: Three capacitors 1μF, 4μF & 7μF give maximum capacity, when connected- (A) All in series (B) All in parallel (C) 1μF & 4μF in series and 7μF in parallel (D) 4μF & 7μF in parallel & 1μF in series

Answer: (B) All in parallel

1. The equivalent capacitance is maximum when the capacitors are connected in parallel.
2. In parallel, C = 1 + 4 + 7 = 12 μF, which is the largest possible value.
3. Hence, the answer is (B).

Teacher's Note:
a) Parallel combinations always give the largest capacitance.
b) Series combinations are always smaller than the smallest capacitor.

Ex.29: Capacitors of 10μF, 25μF, 115μF, 108μF, are connected in parallel, the resultant capacity will be- (A) less than 10μF (B) more than 115μF (C) average of all the four (D) none of these

Answer: (B) more than 115μF

1. In parallel, C = 10 + 25 + 115 + 108 = 258 μF.
2. This is more than the largest capacitor, 115 μF.
3. Hence, the answer is (B).

Teacher's Note:
a) The resultant of a parallel combination is always greater than the largest capacitor.
b) Add the capacitances directly.

Ex.30: Capacitor C, 2C, 4C ... ∞ are connected in parallel, their effective capacitance will be- (A) C (B) C/2 (C) ∞ (D) 0

Answer: (C) ∞

1. C' = C + 2C + 4C + ...... = C(1 + 2 + 4 + ......) = ∞.
2. Hence, the answer is (C).

Teacher's Note:
a) The sum of an infinite increasing series diverges.
b) Parallel capacitances always add.

Ex.31: A capacitor of 20μF is charged to 500 volt, which is then connected in parallel to 10μF capacitor of 200 volt. What will be their common potential?

Answer: 400 volt

1. Charge on the first capacitor: q₁ = C₁V₁ = 20 × 10^-6 × 500 = 0.01 C.
2. Charge on the second capacitor: q₂ = C₂V₂ = 10 × 10^-6 × 200 = 0.002 C.
3. Total charge q = 0.01 + 0.002 = 0.012 C.
4. Total capacitance C = 20 × 10^-6 + 10 × 10^-6 = 30 × 10^-6 F.
5. Common potential V = q / C = 0.012 / (30 × 10^-6) = 400 V.

Teacher's Note:
a) When capacitors are connected in parallel, the capacitances add but energy is not conserved.
b) Total charge is conserved on connection.

Ex.32: In the given fig. the resultant capacity will be- (A) n(n + 1)C (B) n(n² + 1)C (C) nC / (n² + 1) (D) n²C / (n² + 1)

[Figure: Two groups of capacitors between points A and B. The left group has n capacitors of capacity C connected in series. The right group has n capacitors of capacity C connected in parallel, joined to the series group through a further network of capacitors, with the resultant measured between A and B.]

Answer: (C) nC / (n² + 1)

1. The first n capacitors in series give C₁ = C / n.
2. The second n capacitors in parallel give C₂ = nC.
3. C₁ and C₂ are in series: 1/C' = 1/C₁ + 1/C₂ = n/C + 1/(nC) = (n² + 1) / (nC).
4. Hence, C' = nC / (n² + 1), which is option (C).

Teacher's Note:
a) In series, the charge is the same on each capacitor, and the voltage is shared.
b) Use the reciprocal form before inverting the final result.

Ex.33: Seven capacitors each of capacitance 2μF are to be connected in a configuration to obtain an effective capacitance of 10/11 μF. Which of the combination(s) shown in fig will achieve the desired result?

[Figure: Four circuit diagrams (A), (B), (C) and (D), each showing seven 2 μF capacitors arranged in combinations of series and parallel branches between two terminals.]

Combinations in the figure: (A) five capacitors in parallel, in series with two capacitors; (B) four in parallel, in series with three; (C) three in parallel, in series with four; (D) two in parallel, in series with five (all 2 μF).

Answer: (A)

1. In combination (A), as worked in the printed solution: 1/C = 1/(5 × 2) + 2/2 = 1/10 + 1 = 11/10.
2. So C = 10/11 μF.
3. Hence, combination (A) gives the desired capacitance.

Teacher's Note:
a) Break a network into parallel branches, then find each branch as a series combination.
b) Check each option separately; only (A) gives 10/11 μF.

Ex.34: A parallel plate capacitor with air between the plates has a capacitance of 8pF. The capacitance, if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 will be- (A) 96 pF (B) 96 μF (C) 12 μF (D) 12pF

Answer: (A) 96 pF

1. For air: C = ε₀A / d = 8 pF.
2. With d → d/2 and K → 6: C' = 6 × ε₀A / (d/2) = 12 × ε₀A / d.
3. C' = 12 × 8 pF = 96 pF.
4. Hence, the answer is (A).

Teacher's Note:
a) Halving the distance doubles the capacitance, and the dielectric multiplies it by 6.
b) Both effects multiply, giving a factor of 12.

Ex.35: For making a parallel-plate capacitor you have available two plates of copper, which is most appropriate dielectric (A) A sheet of mica (thickness = 0.10 mm K = 5.4) (B) A sheet of glass (thickness = 0.20 mm K = 7) (C) A sheet of paraffin ⅓ (thickness = 1 cm K = 2½) (D) None of the above can be used

Answer: (A) A sheet of mica

1. Capacitance C = K ε₀A / d, so for maximum C the ratio K / d must be maximum.
2. For mica, K / d = 5.4 / 0.1 = 54.
3. For glass, K / d = 7 / 0.2 = 35.
4. For paraffin, K / d = 2.5 / 10 = 0.25 (thickness 1 cm = 10 mm).
5. Hence, mica gives the maximum capacitance, and the answer is (A).

Teacher's Note:
a) Use the same units of thickness before comparing K / d.
b) Thin, high-K dielectrics give the largest capacitance.

Ex.36: A parallel plate capacitor has capacitance 'C' in air. When it is dipped completely in liquid of constant 80, its capacity will becomes- (A) 80 times (B) 1/80 times (C) Remains unchanged (D) 0

Answer: (A) 80 times

1. In air, C = ε₀A / d.
2. In the liquid of K = 80, C' = 80 ε₀A / d = 80C.
3. Hence, the capacity becomes 80 times.

Teacher's Note:
a) A dielectric always increases capacitance by the factor K.
b) Complete immersion means the full area and gap are filled.

Ex.37: If C is the capacitance of a capacitor in air. When dielectric of same thickness and of constants K₁, K₂ & K₃ are completely filled along the plates, then the new capacity will become- (A) 3C K₁K₂K₃ / (K₂K₃ + K₃K₁ + K₁K₂) (B) C/3 (K₁K₂K₃) / (K₂K₃ + K₃K₁ + K₁K₂) (C) 3C K₁K₂K₃ / (K₂K₃ + K₃K₁ + K₁K₂) (D) C/3 K₁K₂K₃ / (K₁K₂ + K₂K₃ + K₃K₁)

Answer: (C) 3C K₁K₂K₃ / (K₂K₃ + K₃K₁ + K₁K₂)

1. In air, with plate separation 3t (three slabs of thickness t stacked across the gap), C = ε₀A / 3t.
2. The three dielectric layers are in series across the gap, so C' = ε₀A / (t/K₁ + t/K₂ + t/K₃).
3. Substituting, C' = ε₀A K₁K₂K₃ / (t(K₂K₃ + K₃K₁ + K₁K₂)).
4. Using C = ε₀A / 3t, C' = 3C K₁K₂K₃ / (K₂K₃ + K₃K₁ + K₁K₂), which is option (C).

Teacher's Note:
a) Slabs stacked across the gap are in series, so their reciprocal dielectric terms add.
b) Slabs covering the full area side by side are in parallel, a different case.

Ex.38: In the above example (41), if the dielectric is filled perpendicular to plates, and the areas of plates in contact with same dielectric are same, then the capacitance will be- (A) 3C (K₁ + K₂ + K₃) (B) C/3 (K₁ + K₂ + K₃) (C) 3C (K₁K₂ + K₂K₃ + K₃K₁) (D) C/3 (1/K₁ + 1/K₂ + 1/K₃)

Answer: (B) C/3 (K₁ + K₂ + K₃)

1. The plate area A is divided into three equal parts A/3, each filled with one dielectric and spanning the full gap d.
2. Each part is a separate parallel-plate capacitor: C_i = K_iε₀(A/3) / d = K_i C / 3, where C = ε₀A / d.
3. The three parts are in parallel, so C' = (C / 3)(K₁ + K₂ + K₃).
4. Hence, the answer is (B). The printed key shows (D), which is the series form and does not match this working.

Teacher's Note:
a) Side-by-side dielectrics across the full gap are in parallel, so their capacitances add.
b) The option (D) form applies to layers stacked across the gap, which is the case in Ex.37.

Ex.39: If the capacity of a parallel plate capacities in air is 1μF and 80μF in an another medium. The dielectric constant of medium will be- (A) 80 (B) 1/80 (C) 40 (D) 1/40

Answer: (A) 80

1. K = C_medium / C_air = (80 × 10^-6) / (1 × 10^-6) = 80.
2. Hence, the dielectric constant is 80.

Teacher's Note:
a) The dielectric constant is the ratio of the two capacitances with the same geometry.
b) Dimensionless quantities do not need units.

Ex.40: A capacitor has two circular plates, whose radius are 8 cm and distance between them is 1mm. When mica (dielectric constant = 6) is placed between the plates, find the capacitance of this capacitor & the energy stored, when it is given potential of 15 volt.

Answer:
1. Radius R = 8 cm = 8 × 10^-2 m, so area A = πR² = π × (8 × 10^-2)² = 0.0201 m².
2. Separation d = 1 mm = 1 × 10^-3 m.
3. Capacitance C = Kε₀A / d = 6 × 8.85 × 10^-12 × 0.0201 / (1 × 10^-3) = 1.068 × 10^-9 F.
4. Energy stored at V = 15 V: U = ½CV² = ½ × 1.068 × 10^-9 × (15)² = 1.2 × 10^-7 J.

Teacher's Note:
a) The printed working takes the potential as 150 V, which is wrong; the question gives 15 V.
b) With the correct 15 V, the energy is 1.2 × 10^-7 J, not 1.2 × 10^-5 J.
c) Use the area of a circle, πR², with the radius in metres.

Ex.41: A parallel plate capacitor has potential 20kV and capacitance 2 × 10^-4 μF. If area of plate be 0.01 m² and distance between the plates be 2mm then find- (A) total electric flux (B) total potential gradient (C) dielectric constant of medium (D) energy

Answer:
1. Given V = 20000 V, A = 0.01 m², d = 0.002 m, C = 2 × 10^-4 μF = 2 × 10^-10 F.
2. (A) Total electric flux = Q / ε₀ = CV / ε₀ = (2 × 10^-10 × 2 × 10⁴) / (8.85 × 10^-12) = 4.52 × 10⁵ V m.
3. (B) Potential gradient = V / d = 20000 / 0.002 = 10⁷ V/m.
4. (C) From C = Kε₀A / d, K = Cd / (ε₀A) = (2 × 10^-10 × 0.002) / (8.85 × 10^-12 × 0.01) = 4.52.
5. (D) Energy U = ½CV² = ½ × 2 × 10^-10 × (2 × 10⁴)² = 4 × 10^-2 J.

Teacher's Note:
a) Total flux through the plates is charge divided by ε₀.
b) Potential gradient is the field strength, V / d.
c) Work out the constant K from the capacitance formula with the given area and separation.

Ex.42: A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in the process of inserting the slab, then (A) Q = ε₀AV / d (B) Q = ε₀KAV / d (C) E = V / Kd (D) W = ε₀AV² / 2d (1 - 1/K)

Answer: (A), (C) and (D)

1. Before insertion, C = ε₀A / d and Q = CV = ε₀AV / d. Since the battery is disconnected, Q stays the same, so (A) is correct.
2. After insertion, C' = KC, and the field is E = V' / d where V' = Q / C' = V / K, so E = V / Kd, which is (C).
3. Energy before: U_i = ½CV². Energy after: U_f = Q² / (2KC) = CV² / (2K).
4. Work done on the system in the process: W = U_f - U_i = ½CV²(1/K - 1) = (ε₀AV² / 2d)(1/K - 1), which matches option (D) in the printed working.

Teacher's Note:
a) With the battery removed, charge is constant, so the potential and energy change.
b) Option (B) wrongly multiplies the charge by K.
c) The energy after insertion is lower, so the field does work on the slab; the work term is negative.

Ex.43: A slab X is placed between the two parallel isolated charged plates, as shown. If Ep and Eq denote the intensity of electric field at P and Q- (A) Ep is reduced by the presence of X if X is metallic (B) Eq is increased by the presence of X if X is dielectric (C) Eq is in the opposite sense to Ep if X is a dielectric. (D) Eq is zero if X is metallic

[Figure: Two large parallel isolated charged plates, the left one positive (marked +) and the right one negative (marked -). A slab X is placed between them. Point P is marked near the positive plate and point Q is marked inside the slab X.]

Answer: (A), (C) and (D)

1. A dielectric placed in an external field becomes polarised, and the induced field opposes the external field inside the dielectric. So E_q is in the opposite sense to E_p when X is a dielectric, which is (C).
2. Inside a metal, the free charges rearrange until the field inside is zero, so E_q = 0 when X is metallic, which is (D).
3. The induced charges on a metallic slab reduce the field at P outside it, so E_p is reduced when X is metallic, which is (A).
4. A dielectric reduces the field inside it, so (B) is not correct.

Teacher's Note:
a) The resultant field inside a dielectric is E / K, which is reduced by the induced field.
b) For a metal, K tends to infinity, so the field inside it is zero.
c) The printed working does not show a separate key line; the answer follows from the reasoning given there.

Ex.44: Half of the space between parallel plate capacitor is filled with a medium of dielectric constant K parallel to the plates. If initially the capacity was C, the new capacity will be-

Answer: C' = 2CK / (1 + K)

1. Each half of the gap is a separate layer: one of air with thickness d/2 and capacitance C₁ = ε₀A / (d/2) = 2C.
2. The other layer has dielectric K and thickness d/2, so C₂ = Kε₀A / (d/2) = 2KC.
3. These two layers are in series: C' = C₁C₂ / (C₁ + C₂) = (2C × 2KC) / (2C + 2KC) = 2CK / (1 + K).

Teacher's Note:
a) Layers placed between the plates one after another are in series.
b) For K = 1, the formula gives C, as expected for air alone.

7. CAPACITANCE FOR DIFFERENT GEOMETRICAL FIGURES

7.1 (i) Parallel plate capacitor
C = ε₀KA / d, where A is the area of each plate, d the distance between the plates, and K the dielectric constant of the medium between the plates. For air, K = 1 and C = ε₀A / d.
(ii) When a number of mediums are placed between the plates, C = ε₀A / (t₁/K₁ + t₂/K₂ + t₃/K₃), where t₁ + t₂ + t₃ = d.
(iii) In this combination (slabs stacked across the gap), the equivalent capacitance is C = (ε₀/d)(K₁A₁ + K₂A₂ + K₃A₃) when the slabs fill the area side by side.
(iv) If a metal sheet of thickness t is kept between the two plates, then C = ε₀A / (d - t).
(v) If a thin metal foil is placed between the plates, then the capacity remains unaffected.
(vi) C_medium / C_air = K.
Note: A parallel plate capacitor is charged and the battery is removed. Distance reduced: charge stays same, capacity increases, potential difference decreases, field does not change, energy decreases. Dielectric placed between plates: charge stays same, capacity increases, potential difference decreases, field decreases, energy decreases. A parallel plate capacitor kept connected to a battery: distance reduced gives same potential difference, increased capacity, increased charge, increased field and increased energy; dielectric placed gives same potential difference, increased capacitance, increased charge, no change in field, and increased energy.
(vii) Condensers provide infinite resistance for a D.C. circuit and allow A.C. to pass through them.

For the slabs of thickness t₁, t₂, t₃ placed one after another between the plates, \( C = \frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}} \). When the dielectrics K₁, K₂, K₃ fill areas A₁, A₂, A₃ side by side, the equivalent capacitance is \( C = \frac{K_1\varepsilon_0A_1}{d}+\frac{K_2\varepsilon_0A_2}{d}+\frac{K_3\varepsilon_0A_3}{d} = \frac{\varepsilon_0}{d}(K_1A_1+K_2A_2+K_3A_3) \).

7.2 Spherical condenser

(A) When the outer sphere is earthed: \( C = 4\pi\varepsilon_0\frac{R_1R_2}{R_2-R_1} \).

(B) When the inner sphere is earthed: this situation can be visualized as a parallel combination of two capacitors, one between the two spheres and the second between the outer sphere and infinity. Therefore \( C = \frac{4\pi\varepsilon_0K_1R_1R_2}{R_2-R_1} + 4\pi\varepsilon_0K_2R_2 \). If K₁ = K₂ = K then \( C = \frac{4\pi\varepsilon_0KR_2^2}{R_2-R_1} \).

If charge Q is given to the outer sphere, it is divided partly between its inner and outer surfaces. Let the charge on the inner sphere be q'; the potential of the inner sphere is 0. Charge on the inner surface of the outer sphere = q = \( \frac{R_1}{R_2}Q \). Charge on the outer sphere = (Q – q) = \( \frac{R_2-R_1}{R_2}Q \). Charge on the inner sphere = q' = \( -\frac{R_1}{R_2}Q \).

7.3 Capacitance of cylindrical capacitor

If R₁ is the radius of the inner cylinder, R₂ the radius of the outer cylinder and ℓ the length of the cylinder, then \( C = \frac{2\pi\varepsilon_0K\ell}{\ln\frac{R_2}{R_1}} = \frac{2\pi\varepsilon_0K\ell}{2.303\log_{10}\left(\frac{R_2}{R_1}\right)} \).

Ex.45: A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32. Determine- (i) The capacitance of the capacitor and (ii) Potential of the inner sphere

Answer:
1. Radii R₁ = 12 cm = 0.12 m, R₂ = 13 cm = 0.13 m, K = 32.
2. (i) C = 4πε₀K R₁R₂ / (R₂ - R₁) = (1 / 9 × 10⁹) × 32 × (12 × 13 × 10^-4) / (10^-2) = 5.55 × 10^-9 F.
3. (ii) Potential of inner sphere: V = q / C = (2.5 × 10^-6) / (5.55 × 10^-9) = 4.5 × 10² V.

Teacher's Note:
a) Use the radii in metres, not centimetres.
b) The liquid between the spheres multiplies the capacitance by K.
c) Potential of the inner sphere is charge divided by capacitance.

Ex.46: A cylindrical capacitor has two co-axial cylinders of 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 mC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e. bending of filled lives at the ends)

[Figure: Two coaxial cylinders of radii 1.4 cm (inner) and 1.5 cm (outer), with a length of 15 cm. The inner cylinder is marked R₁ and the outer cylinder R₂.]

Answer:
1. Length ℓ = 15 cm = 0.15 m, inner radius a = 1.4 cm = 0.014 m, outer radius b = 1.5 cm = 0.015 m.
2. Capacitance C = 2πε₀ℓ / (2.303 log₁₀(b / a)) = (2π × 8.854 × 10^-12 × 0.15) / (2.303 × log₁₀(1.5 / 1.4)) = 1.21 × 10^-10 F.
3. Potential of inner cylinder: V = q / C = (3.5 × 10^-6) / (1.21 × 10^-10) = 2.89 × 10⁴ V, when the charge is taken as 3.5 μC.

Teacher's Note:
a) The printed charge is 3.5 mC, but a 3.5 mC charge gives 2.89 × 10⁷ V, which is not practical; the printed working uses 3.5 μC.
b) The logarithm uses the ratio of outer to inner radius, so keep the same units for both.
c) Keep 2.303 log₁₀ in the denominator, which equals ln.

Ex.47: A capacitor is discharged through a 10 MΩ resistor and it is found that the time constant is 200s. Then, find the value of the capacitance.

Answer:
1. Time constant τ = RC = 200 s.
2. C = τ / R = 200 / (10 × 10⁶) = 20 × 10^-6 F = 20 μF.

Teacher's Note:
a) Convert MΩ to ohms before dividing.
b) The time constant is the product of resistance and capacitance.

Ex.48: A 2500 μF capacitor is charged through a 1 KΩ resistor by a 12V d.c. source. What is the voltage across the capacitor after 5 sec?

Answer:
1. Time constant τ = RC = 10³ × 2500 × 10^-6 = 2.5 s.
2. Charging: V = V₀(1 - e^-t/RC).
3. With t = 5 s, t / RC = 2, so V = 12(1 - e^-2) = 12 × 0.865 = 10.38 V.

Teacher's Note:
a) Compute t / RC first, here 5 / 2.5 = 2.
b) After five time constants the capacitor is almost fully charged.

Ex.49: Capacitor C₁ of capacitance 1 microfarad and capacitor C₂ of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. (A) The current in each of the two discharging circuits is zero at t = 0 (B) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C₁ loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge.

Answer: (B) and (D)

1. Both capacitors are charged to the same voltage V by the same battery, and the resistors are equal.
2. Initial discharge current I₀ = V / R for both, which is the same and non-zero, so (A) and (C) are not correct.
3. For discharge, q = q₀e^-t/RC. Half the charge is lost when t = RC ln 2.
4. So τ₁ / τ₂ = (RC₁) / (RC₂) = C₁ / C₂ = 1/2, which means C₁ loses 50% of its charge sooner than C₂.
5. Hence, the correct statements are (B) and (D).

Teacher's Note:
a) The initial current V / R is the same for both circuits, because both capacitors start at the same voltage.
b) The smaller capacitor discharges faster through the same resistor.
c) Statement (B) holds at t = 0, where both currents equal V / R and are not zero.

Ex.50: In the adjoining diagram Fig, the condenser C will be fully charged to potential V if- (A) S₁ and S₂ both are open (B) S₁ and S₂ both are closed (C) S₁ is closed and S₂ is open (D) S₁ is open and S₂ is closed.

[Figure: A battery V in one branch, with a 5 Ω resistor and switch S₁ in series, a switch S₂ with a 10 Ω resistor in another branch, and a capacitor C connected across the circuit.]

Answer: (C) S₁ is closed and S₂ is open

1. For the capacitor to charge, the battery must be connected to it through a path with S₁ closed.
2. If S₂ is closed, the capacitor would discharge through the other branch, so S₂ must be open.
3. Hence, the condenser is fully charged when S₁ is closed and S₂ is open, which is option (C).

Teacher's Note:
a) Both switches closed would charge and discharge the capacitor at the same time.
b) Open switches leave the capacitor isolated, so it cannot charge.

8. CHARGE AND DISCHARGE OF A CONDENSER

Charge of a condenser
(1) The capacitor takes some finite time in the charging process.
(2) The quantity of charge at any instant of time t is given by q = q₀[1 - e^-t/RC], i.e. the charge rises exponentially. Here q₀ = maximum final value of charge at t = ∞, R = resistance of the whole circuit.
(3) If t = RC = τ (time constant), then q = q₀[1 - 1/e] = 0.63 q₀.
(4) Time constant τ = RC is the time during which the charge rises on the condenser plates to 63% of its maximum value during charging, or falls to 37% of its maximum value during discharge.
(5) The potential difference across the condenser plates at any instant is given by V = V₀[1 - e^-t/RC] volt.
(6) The transient current in the circuit at any instant of time is I = I₀e^-t/RC, where I₀ = V₀/R is the initial current.
(7) If t = RC = τ = time constant, then I = I₀ / e = 0.37 I₀ = 37% of I₀.

Discharge of a condenser
(1) The capacitor takes some finite time in the discharging process.
(2) The quantity of charge on the condenser at any instant of time is given by q = q₀e^-t/RC, i.e. the charge falls exponentially.
(3) If t = RC = τ, then q = q₀ / e = 37% q₀. The time constant is the time during which the charge on the condenser plates during discharge falls to 37% of its maximum value.
(4) The dimensions of RC are those of time, i.e. M⁰L⁰T¹, and dimension of 1/RC are those of frequency, i.e. M⁰L⁰T^-1.
(5) The potential difference across the condenser at any instant is V = V₀e^-t/RC volt.
(6) The transient current decreases exponentially, I = -I₀e^-t/RC; the current is opposite to that of charging.

Charging: q rises from 0 towards q₀; at t = RC = τ, q = 0.63 q₀.

Discharging: q falls from q₀; at t = RC = τ, q = q₀/e = 37% of q₀.

Current–time graph: during charge the current decays from I₀; during discharge it is of opposite direction, from –I₀ towards zero.

Points to remember

1. If n identical condensers, each of capacity C are connected in series, then the effective capacitance of the system is C/n.
2. If n identical capacitors each of capacity C are connected in parallel, the effective capacitance of the system is nC.
3. Energy stored in a condenser or conductor lies in its electric field.
4. Total energy stored in series combination is U = U₁ + U₂ + U₃ + ........ = ½q²/C₁ + ½q²/C₂ + ...
5. The energy stored in parallel combination of condensers is U = U₁ + U₂ + U₃ + ..... = ½C₁V² + ½C₂V² + ...
6. Energy stored per unit volume in the medium between the plates of a parallel plate capacitor is u = ½ε₀K²E², where u is called the energy density.
7. Force of attraction between the plates of a parallel plate capacitor is F = q² / (2ε₀A).
8. Dielectric polarisation: the phenomenon of displacement of centres of positive and negative charges of a non-polar dielectric in the presence of external field is called dielectric polarisation.
9. The induced dipole moment P is proportional to the electric field E.
10. P = αε₀E, where α is a proportionality constant known as atomic polarisability. In scalar form, P = αε₀E.
11. The unit of α is m³.
12. Induced charge on the faces of dielectric due to dielectric polarisation is q' = q(1 - 1/K).
13. If n small drops, each having a charge q, capacity C and potential V coalesce to a big drop, then (a) the charge on the big drop = nq, (b) capacity of big drop = n^1/3C, (c) potential of big drop = n^2/3V, (d) potential energy of big drop = n^5/3U.
14. If dielectric mediums of constant K₁ and K₂ are filled in between the plates of the condenser as shown in the figure, the arrangement is equivalent to two condensers in parallel, hence the effective capacitance of the system becomes C = (ε₀A / 2d)(K₁ + K₂).
15. Equivalent dielectric constant of the medium in the above case is K = (K₁ + K₂) / 2.
16. When the different dielectrics are filled parallel between the plates, the effective dielectric is given by 1/K = 1/K₁ + 1/K₂ + 1/K₃.
17. When the different dielectrics are filled perpendicular between the plates, the effective dielectric is given by K = K₁ + K₂ + K₃.
18. Since q = CV, voltage is proportional to the charge and not to the current.
19. There can be a potential difference across the terminals of the condensers even if there is no current in the circuit.
20. For D.C. circuits the condenser acts as an open circuit because for D.C. circuits current i = 0.
21. If the dielectric medium of constant K₁ and K₂ are filled between the plates of the condenser as shown in the figure, the arrangement is equivalent to two condenser in series. Then the effective capacity is 1/C = 1/C₁ + 1/C₂.
22. A parallel plate capacitor is charged to V volts and charging battery is removed. A dielectric plate is now introduced between the plates of condenser, then (a) charge remains the same, (b) capacity increases K times the previous value, (c) potential difference decreases and becomes 1/K times the previous value, (d) energy decreases and becomes 1/K times the previous value.
23. If the air capacitor is charged to V volt and a dielectric plate is introduced in the gap between the plates (battery still connected), then (a) the charge on the plates increases, (b) the capacity increases K times, (c) the potential difference between the plates remains the same, (d) the energy increases; becomes K times the previous value i.e. U₂ = KU₁.
24. Kirchhoff's law can be applied to the capacitor circuits in a similar manner to the current circuits. It states that the algebraic sum of charges at a junction in the circuit is zero, Σq = 0.

Figures for the points above:

Point 16 (dielectrics K₁, K₂, K₃ of thicknesses t₁, t₂, t₃ stacked one after another between the plates):

Point 17 (dielectrics K₁, K₂, K₃ filling areas A₁, A₂, A₃ side by side between the plates):

(vi) If potential of a conductor rises by 1 volt as a result of charging by 1 coulomb , then capacitance of that conductor is said to be 1 Farad.

Sol. [C =Q/V=Q²/VQ=Q²/J

and C =Q/V =QV/v₂= J/v²

Ans.[D]

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