CBSE Class 12 Physics Electrostatics Solved Examples

Question. F_g and F_e represents gravitational and electrostatic force respectively between electrons situated at a distance 10 cm. The ratio of F_g/ F_e is of the order of:
a. 10⁴²
b. 10
c. 1
d. 43^-10
Answer : D

Question. Which of the following charge is not possible?
a. 18 1.6 10^-18 C 
b. 1.6 10^-19 C
c. 1.6 10^-20 C
d. None of these
Answer : C

Question. A total charge Q is broken in two parts Q₁ and Q₂ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when:

Answer : D

Question. If the radius of a solid and hollow copper spheres are same which one can hold greater charge?
a. Solid sphere
b. Hollow sphere
c. Both will hold equal charge
d. None of these
Answer : C

Question. Number of electrons in one coulomb of charge will be:
a. 5.46 × 10²⁹
b. 6.25 ×10¹⁸
c. 1.6 ×10¹⁹
d. 9 × 10¹¹
Answer : B

Question. The force between two charges 0.06m apart is 5 N. If each charge is moved towards the other by 0.01m, then the force between them will become:
a. 7.20 N
b. 11.25 N
c. 22.50 N
d. 45.00 N
Answer : B

Question. Two point charges 1μC and 5μC are separated by a certain distance. What will be ratio of forces acting on these two?
a. 1 : 5
b. 5 :1
c. 1 : 1
d. 0
Answer : C

Question. Two point charges +3μC and +8μC repel each other with a force of 40 N. If a charge of −5μC is added to each of them, then the force between them will become:
a. −10N
b. +10N
c. +20N
d. −20N
Answer : A

Question. The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5 x 10^-11m will be: (Charge on electron = 1.6 × 10^–19 C, mass of electron = 9.1 × 10^–31 kg, mass of proton = 1.6 10^-27 kg G = 6.7 x 10^-11 Nm²/g²)
a. 2.36 × 1039
b. 2.36 × 1040
c. 2.34 × 1041
d. 2.34 × 1042
Answer : A

Question. Two charges of 40 μC and –20μC are placed at a certain distance apart. They are touched and kept at the same distance. The ratio of the initial to the final force between them is:
a. 8 :1
b. 4 :1
c. 1 : 8
d. 1 : 1
Answer : A

Question. Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of q/m is:

Answer : D

Question. Five balls numbered 1 to 5 balls suspended using separate threads. Pair (1,2), (2,4) and (4,1) show electrostatic attraction, while pair (2,3) and (4,5) show repulsion. Therefore ball 1 must be:
a. Positively charged
b. Negatively charged
c. Neutral
d. Made of metal
Answer : C

Question. ABC is a right angle triangle in which AB = 3 cm, BC = 4 cm and ∠ABC = π/∠ = . The three charges +15, +12 and – 20 e.s.u. are placed respectively on A, B and C. The force acting on B is:

a. 125 dynes
b. 35 dynes
c. 25 dynes
d. Zero
Answer : C

SOLVED EXAMPLES

Ex.1 Two identical conducting small spheres are given charges of opposite sign. These attract each other with a force of 0.3 Newton when placed at a distance of 30 cm. These spheres are then connected by a conducting wire. When the conducting wire us removed, the spheres repel each other with a force of 0.1 Newton at the same distance of 30 cm. Find the initial charges on each sphere.

Sol.:
1. Let the charge on the spheres be +q1 and -q2.
2. Coulomb force of attraction: F = \( \frac{q_1 q_2}{4\pi\epsilon_0 (0.3)^2} \) = 0.3 N
3. q1q2 = \( \frac{0.3 \times (0.3)^2}{9 \times 10^{9}} \) = 3 × 10^-12 C²
4. When the spheres are connected by a conducting wire, charge flows until their potentials are equal. The spheres are identical, so each gets half of the net charge, that is \( \frac{q_1 - q_2}{2} \).
5. Coulomb force after disconnection: \( \frac{\left( \frac{q_1 - q_2}{2} \right)^2}{4\pi\epsilon_0 (0.3)^2} \) = 0.1 N
6. (q1 - q2)² = \( \frac{0.1 \times (0.3)^2 \times 4}{9 \times 10^{9}} \) = 4 × 10^-12
7. q1 - q2 = ±2 × 10^-6 ....(i)
8. q1 + q2 = \( \sqrt{(q_1 - q_2)^2 + 4q_1q_2} \) = \( \sqrt{4 \times 10^{-12} + 4 \times 3 \times 10^{-12}} \) = ±4 × 10^-6 ....(ii)
9. From (i) and (ii): q1 = ±3 μC and q2 = ∓1 μC.
10. The initial charges are +3 μC and -1 μC.

Teacher's Note:
a) Connecting identical spheres shares the net charge equally between them.
b) The sum and difference of the two charges come from squaring and adding the two equations.
c) Attraction gives the product q1q2, and the repulsion after contact gives the difference.

Ex.2 Four charges +2μC, -4μC, +6μC and +4μC are lying at the corners A, B, C and D respectively of a square ABCD of side 0.1 m as shown in the figure. Find the work done in assembling these charges.

Sol.:
1. The four charges are brought from infinity to the corners one by one. The work done equals the sum of the potential energies of all pairs.
2. Total potential energy U = U_AB + U_BC + U_CD + U_DA + U_AC + U_BD
3. U = \( \frac{1}{4\pi\epsilon_0 r} \) [ (2 × 10^-6)(-4 × 10^-6) / 0.1 + (-4 × 10^-6)(6 × 10^-6) / 0.1 + (6 × 10^-6)(4 × 10^-6) / 0.1 + (4 × 10^-6)(2 × 10^-6) / 0.1 + (2 × 10^-6)(6 × 10^-6) / (√2 × 0.1) + (-4 × 10^-6)(4 × 10^-6) / (√2 × 0.1) ]
4. U = \( \frac{10^{-12}}{4\pi\epsilon_0 \times 0.1} \) [ -8 - 24 + 24 + 8 + \( \frac{12}{\sqrt{2}} - \frac{16}{\sqrt{2}} \) ]
5. U = \( \frac{10^{-12} \times 9 \times 10^{9}}{0.1} \) × \( \left[ -\frac{4}{\sqrt{2}} \right] \) = -\( \frac{36}{100\sqrt{2}} \)
6. U = -0.254 J
7. The work done in assembling the charges is -0.254 J.

Teacher's Note:
a) Work of assembly equals the total electrostatic energy of all pairs of charges.
b) The two diagonal pairs use the diagonal length √2 × 0.1 m, not 0.1 m.
c) The negative result means the system is more stable than with the charges at infinity, since the attractive pairs dominate.

Ex.3 A small sphere of mass 100 mg is given a charge of +2.5 × 10^-10 coulomb. It hangs from an insulating and inextensible string 0.2 m long. The upper end of the string is connected to a large positively charged plane vertical conducting plate which has a surface charge density σ = 1.5 × 10^-5 C.m^-2. Find the angle which the string makes with the vertical direction at equilibrium.

Sol.:
1. The electric field due to the large conducting sheet: E = \( \frac{\sigma}{\epsilon_0} \)
2. Repulsive force on the charge q on the sphere: F = qE = \( \frac{q\sigma}{\epsilon_0} \)
3. Forces on the sphere: weight mg (downwards), electrostatic repulsion F (horizontal), and tension T along the string.
4. For equilibrium: sum of F_x = 0, so F = T sin θ ....(iv)
5. Sum of F_y = 0, so mg = T cos θ ....(v)
6. Dividing (iv) by (v): tan θ = \( \frac{F}{mg} = \frac{q\sigma}{\epsilon_0 mg} \)
7. tan θ = \( \frac{(2.5 \times 10^{-10}) \times (1.5 \times 10^{-5})}{(8.85 \times 10^{-12}) \times (100 \times 10^{-6}) \times 9.8} \) = 0.432
8. θ = tan^-1 (0.432) = 23.4°

Teacher's Note:
a) The field of a conducting plate is σ/ε0, not σ/2ε0, because the field is measured at a conductor's surface.
b) Take the vertical and horizontal components of the tension separately.
c) Convert the mass from milligrams to kilograms before using g = 9.8 m/s².

Ex.4 A thin vertical wire ring of radius R and a very long uniformly charged straight wire are so placed that one end of the wire coincides with the centre of the ring. The total charge on the ring is Q and the wire has a uniform charge per unit length equal to λ. Find the interaction force between the ring and the straight wire.

Sol.:
1. Electric field at P on the axis of the ring: E = \( \frac{1}{4\pi\epsilon_0} \frac{Qx}{(R^2 + x^2)^{3/2}} \)
2. Charge on an element dx of the wire is λ dx, so the force on this element is dF = \( \frac{1}{4\pi\epsilon_0} \frac{Q \cdot \lambda dx \cdot x}{(R^2 + x^2)^{3/2}} \) = \( \frac{Q\lambda x \, dx}{4\pi\epsilon_0 (R^2 + x^2)^{3/2}} \)
3. Total force: F = \( \int_0^{\infty} dF = \frac{Q\lambda}{4\pi\epsilon_0} \int_0^{\infty} \frac{x \, dx}{(R^2 + x^2)^{3/2}} \)
4. Put x = R tan θ, so dx = R sec²θ dθ and R² + x² = R² sec²θ.
5. F = \( \frac{Q\lambda}{4\pi\epsilon_0 R} \int_0^{\pi/2} \sin\theta \, d\theta = \frac{Q\lambda}{4\pi\epsilon_0 R} [-\cos\theta]_0^{\pi/2} = \frac{Q\lambda}{4\pi\epsilon_0 R} \)
6. The force is repulsive for like charges, directed along the axis of the ring, away from the ring.

Teacher's Note:
a) The force on the wire is the field of the ring multiplied by the charge of the element.
b) The substitution x = R tan θ turns the integral into a simple sine integral.
c) The printed final line drops R; the correct result is Qλ/(4πε0 R).

Ex.5 Three concentric spherical shells, made of conducting materials, have radii R1, R2 and R3 and are given charges Q1, Q2 and Q3 respectively as shown in the figure. Find the potential for points at distance r when
(a) r > R1
(b) R1 ≤ r ≤ R2
(c) R2 ≤ r ≤ R3
(d) r > R3

Sol.:
1. The potential at any point within a conducting spherical shell is the same as that at its outer surface. The potential at any point outside a spherical shell is the same as if its charge were concentrated at the centre.
2. (a) For r less than R1 (the point A inside the innermost shell): V_A = \( \frac{Q_1}{4\pi\epsilon_0 R_1} + \frac{Q_2}{4\pi\epsilon_0 R_2} + \frac{Q_3}{4\pi\epsilon_0 R_3} \) = \( \frac{1}{4\pi\epsilon_0} \left[ \frac{Q_1}{R_1} + \frac{Q_2}{R_2} + \frac{Q_3}{R_3} \right] \)
3. (b) For R1 ≤ r ≤ R2: V_B = \( \frac{1}{4\pi\epsilon_0} \left[ \frac{Q_1}{r} + \frac{Q_2}{R_2} + \frac{Q_3}{R_3} \right] \)
4. (c) For R2 ≤ r ≤ R3: V_C = \( \frac{1}{4\pi\epsilon_0} \left[ \frac{Q_1}{r} + \frac{Q_2}{r} + \frac{Q_3}{R_3} \right] \)
5. (d) For r greater than R3, all charges act as if placed at the centre: V_D = \( \frac{1}{4\pi\epsilon_0 r} [Q_1 + Q_2 + Q_3] \)

Teacher's Note:
a) The printed part (a) says r greater than R1, but the key works the region inside the innermost shell, so the answer given is for r less than R1.
b) Inside a conducting shell, its own charge contributes a constant term equal to its value at the surface.
c) Outside all shells the potential is the same as for a single point charge equal to the total charge.

Ex.6 A cone made of an insulating material has a total charge Q spread uniformly over its sloping surface. Calculate the energy required to bring up a small test charge q from infinity to the apex of the cone. The cone has a slant length L.

Sol.:
1. If R is the radius of the cone and θ is the angle of the cone, then sin θ = R / L.
2. Surface charge density: σ = \( \frac{Q}{\pi R L} \)
3. Consider an element of the cone of radius r, lying between slant lengths l and l + dl from the apex A.
4. Area of the element: dA = 2πr dl, and its charge: dq = σ dA = σ(2πr dl)
5. Potential at the apex due to this ring element: dV = \( \frac{2\pi r \, dl \, \sigma}{4\pi\epsilon_0 l} \)
6. Potential at the apex due to all elements: V_A = \( \int_0^L \frac{2\pi r \sigma \, dl}{4\pi\epsilon_0 l} \), with r = l sin θ, so V_A = \( \frac{\sigma}{2\epsilon_0} (L \sin\theta) \)
7. V_A = \( \frac{\sigma}{2\epsilon_0} \times L \times \frac{R}{L} = \frac{\sigma R}{2\epsilon_0} = \frac{QR}{\pi R L \times 2\epsilon_0} = \frac{Q}{2\pi\epsilon_0 L} \)
8. Work done in bringing the charge q from infinity to the apex: W = V_A q = \( \frac{Qq}{2\pi\epsilon_0 L} \)

Teacher's Note:
a) Every ring of the cone is at a different distance from the apex, so the integral is needed.
b) The sine of the cone angle is R/L, which simplifies the result to Q/L.
c) The work equals the charge q times the potential at the apex, and it does not depend on R.

Ex.7. A potential difference of 600 volt is applied across two parallel plates, each of length 1.5 cm, separated by a distance of 5.0 mm. If an electron of energy 1000 eV enters at right angles to the electric field, find the displacement of the electron in the direction perpendicular to the plates. When it come out of space between plates.

Sol.:
1. Electric field between the plates: E = \( \frac{V}{d} = \frac{600}{5 \times 10^{-3}} = 1.2 \times 10^{5} \) V/m
2. Force on the electron: F = eE, so acceleration along the y-axis: a = \( \frac{F}{m} = \frac{eE}{m} \)
3. Time taken to cross the plates along x: t = L / v0, where v0 is the initial velocity of the electron.
4. Displacement along the y-axis: y = \( \frac{1}{2} a t^2 = \frac{1}{2} a \left( \frac{L}{v_0} \right)^2 \) = \( \frac{eEL^2}{2mv_0^2} \)
5. Kinetic energy of the electron: K = \( \frac{1}{2} mv_0^2 \) = 1000 eV = 1000 × 1.6 × 10^-19 J
6. Substituting, y = \( \frac{eEL^2}{4K} \)
7. y = \( \frac{(1.6 \times 10^{-19})(1.2 \times 10^{5})(1.5 \times 10^{-2})^2}{4 \times 1000 \times 1.6 \times 10^{-19}} \) = 6.75 × 10^-3 m = 6.75 mm

Teacher's Note:
a) Convert the plate length to metres (1.5 cm = 0.015 m) before squaring it.
b) The energy is given in electron volts, so convert it to joules with 1 eV = 1.6 × 10^-19 J.
c) The displacement depends on the square of the plate length and on the field, but it falls as the kinetic energy rises.

Ex.8. A parallel plate capacitor has each plate of area 0.02 m² separated by 5.0 mm air gap. The capacitor is connected to a 500 volt supply. Find :
(a) Its capacitance and charge on the plates.
(b) The electric field between the plates and energy stored in the condenser.
(c) The additional charge flowing to the capacitor from the power supply if a liquid of dielectric constant 2.5 is poured between the plates.

Sol.:
1. (a) Capacitance of a parallel plate capacitor: C = \( \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.02}{5 \times 10^{-3}} \) = 3.54 × 10^-11 F
2. Charge on each plate: q = CV = 3.54 × 10^-11 × 500 = 1.77 × 10^-8 C
3. (b) Energy stored: U = \( \frac{1}{2} qV = \frac{1}{2} \times 1.77 \times 10^{-8} \times 500 \) = 4.425 × 10^-6 J
4. Electric field between the plates: E = \( \frac{V}{d} = \frac{500}{5 \times 10^{-3}} \) = 1.0 × 10^5 V/m
5. (c) Capacitance after the dielectric is introduced: C' = 2.5 × 3.54 × 10^-11 = 8.85 × 10^-11 F
6. New charge at the same voltage: q' = C'V = 8.85 × 10^-11 × 500 = 4.425 × 10^-8 C
7. Additional charge from the source: q' - q = 4.425 × 10^-8 - 1.77 × 10^-8 = 2.655 × 10^-8 C

Teacher's Note:
a) Keep the plate separation in metres, 5.0 mm = 5 × 10^-3 m.
b) The energy stored is half of charge times voltage, not the full product.
c) A dielectric of constant K multiplies the capacitance by K, so the extra charge is (K - 1) times the original charge.

Ex.9 A capacitor of capacitance 4μF is charged to a potential difference of 100 V and another of capacitance 6μF is charged to a potential of 200 V. These capacitances are now joined with plates of like charges connected together. Calculate :
(a) the total potential across each after joining
(b) the total electrical energy stored before joining
(c) the total electrical energy stored after joining
How do you account for the difference in energies in (b) and (c) ?

[Figure: Two capacitors drawn side by side. The upper one is labelled 200 V and 6μF, and its positive plate is shown at the top. The lower one is labelled 100 V and 4μF, and its positive plate is shown at the top. A switch S is shown on the left, connecting the two capacitors.]

Sol.:
1. V1 = 100 V, C1 = 4 μF; Q1 = C1V1 = 4 × 10^-6 × 100 = 4 × 10^-4 C
2. V2 = 200 V, C2 = 6 μF; Q2 = C2V2 = 6 × 10^-6 × 200 = 12 × 10^-4 C
3. (a) Total charge: Q = Q1 + Q2 = (4 + 12) × 10^-4 = 1.6 × 10^-3 C
4. Capacitance of the combination in parallel: C = C1 + C2 = (4 + 6) μF = 10 μF
5. Potential across each capacitor after joining: V = \( \frac{Q}{C} = \frac{1.6 \times 10^{-3}}{10 \times 10^{-6}} \) = 160 V
6. (b) Energy before joining: W1 = \( \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (100)^2 \) = 2 × 10^-2 J
7. W2 = \( \frac{1}{2} C_2 V_2^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (200)^2 \) = 12 × 10^-2 J
8. Total energy before joining: W_i = W1 + W2 = (2 + 12) × 10^-2 = 0.14 J
9. (c) Energy after joining: W_f = \( \frac{1}{2} \times 10 \times 10^{-6} \times (160)^2 \) = 0.128 J
10. The energy after joining is less by (0.14 - 0.128) J = 0.012 J. This energy is dissipated as heat in the connecting wires as the current flows.

Teacher's Note:
a) Like plates joined together share the total charge, so the capacitances add in parallel.
b) Energy is found from ½CV² for each capacitor before joining.
c) The loss of energy is not a mistake; it is the heat produced during the charge flow.

Ex.10 A slab of thickness t and dielectric constant K is introduced between the plates of a parallel plate capacitor such that its faces are parallel to the plates and are of same surface area of cross-section as that of plates of the condenser.
(a) Find the capacitance of the capacitor if the separation between the plates is d.
(b) If the dielectric constant is equal to 5.0, find the value of t/d so that the capacitance of the system becomes thrice the capacitance with free space in between the plates.
(c) Find the ratio of energy stored in capacitors in both the cases.

[Figure: Parallel plate capacitor with plate area A marked at the top. The gap between the plates is marked d. A dielectric slab of thickness t and constant K is shown inside the gap, touching the plates, with its thickness t marked at the right.]

Sol.:
1. (a) The dielectric slab acts like two capacitors in series. The area of the plates is the same. The separation is (d - t) for the air capacitor and t for the capacitor with dielectric.
2. Capacity of the air capacitor: C1 = \( \frac{\epsilon_0 A}{d - t} \)
3. Capacity of the capacitor with dielectric: C2 = \( \frac{K \epsilon_0 A}{t} \)
4. For the series combination: \( \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d - t}{\epsilon_0 A} + \frac{t}{K \epsilon_0 A} = \frac{1}{\epsilon_0 A} \left[ (d - t) + \frac{t}{K} \right] \)
5. C = \( \frac{\epsilon_0 A}{(d - t) + \frac{t}{K}} \)
6. (b) Capacitance of the air capacitor: C' = \( \frac{\epsilon_0 A}{d} \)
7. Given C = 3C': \( \frac{\epsilon_0 A}{(d - t) + t/K} = \frac{3\epsilon_0 A}{d} \), so (d - t) + \( \frac{t}{K} \) = \( \frac{d}{3} \)
8. Rearranging: 2d = 3t \( \left[ 1 - \frac{1}{K} \right] = 3t \left[ \frac{K - 1}{K} \right] \)
9. \( \frac{t}{d} = \frac{2K}{3(K - 1)} = \frac{2 \times 5}{3 \times 4} = \frac{5}{6} \) = 0.83
10. (c) Energy is \( \frac{q^2}{2C} \) for the same charge q. So U = \( \frac{q^2}{2C} \) for the dielectric case and U' = \( \frac{q^2}{2C'} \) for the air case.
11. \( \frac{U}{U'} = \frac{C'}{C} = \frac{1}{3} \)

Teacher's Note:
a) A slab that fills part of the gap behaves as two capacitors in series, one of them with air.
b) Keep the ratio t/d below 1, because the slab cannot be thicker than the gap.
c) Energy in the same charge case is inversely proportional to capacitance, so the energy falls when capacitance rises.

Ex.11 A 10μF condenser C is charged through a resistance R of 0.1 MΩ from a battery of 1.5 V. Find the time required for the capacitor to get charged upto 0.75 V for the circuits shown in figure in case (a) and (b) when switch S is switched on.

[Figure: Circuit (a) shows a battery E, a resistance R in the top wire, a switch S in the bottom wire, and a capacitor C on the right side. Circuit (b) shows a switch S at the top, a battery on the left, the capacitor C in the middle branch, and the resistance R on the right side in parallel with C.]

Sol.:
1. (a) For charging of a capacitor C through a resistance R: q = q0 [1 - e^(-t/RC)], so e^(-t/RC) = 1 - \( \frac{q}{q_0} \)
2. For a capacitor, q = CV, so \( \frac{q}{q_0} = \frac{V}{V_0} = \frac{0.75}{1.5} = \frac{1}{2} \)
3. e^(-t/RC) = 1 - \( \frac{1}{2} \) = \( \frac{1}{2} \), so \( \frac{-t}{RC} = -\log_e 2 \)
4. t = RC log_e 2 = (0.1 × 10^6) × (10 × 10^-6) × log_e 2 = log_e 2 = 0.693 second.
5. (b) In this circuit the capacitor C is connected directly to the battery initially, so it acts like a short circuit at first. The capacitor gets charged instantaneously.
6. Therefore t = 0 second.

Teacher's Note:
a) The time constant RC equals 1 second here, because 0.1 MΩ times 10 μF is 1 s.
b) Charging to half the final voltage takes RC ln 2, which is about 0.69 RC.
c) In circuit (b) the resistance does not limit the current, so there is no delay.

Ex.12 A capacitor has square plates of side a making small angle θ with each other. Find the value of its capacitance if d is the minimum separation between plates at one end.

Sol.:
1. The spacing between the plates increases continuously. Consider a strip of width Δx lying at a distance x from the left end.
2. Separation between the plates at the strip: d + x tan θ ≈ d + xθ (since tan θ ≈ θ for small θ)
3. Area of the strip = a Δx
4. Capacitance of the strip: ΔC = \( \frac{\epsilon_0 \, a \, \Delta x}{d + \theta x} \)
5. The capacitance of the capacitor is the sum of the strip capacitances in parallel: C = ∑ΔC. In the limit Δx → 0: C = \( \int_0^{a} \frac{\epsilon_0 a \, dx}{d + \theta x} \)
6. C = \( \frac{\epsilon_0 a}{\theta} \left[ \log_e (d + \theta x) \right]_0^{a} = \frac{\epsilon_0 a}{\theta} \log_e \left( 1 + \frac{\theta a}{d} \right) \)

Teacher's Note:
a) Each thin strip is a small capacitor, and the strips are connected in parallel.
b) The approximation tan θ ≈ θ holds only for small angles.
c) The integral gives a logarithm, not a simple product, because the spacing changes along the plate.

Ex.13 Rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as shown. Two beads of equal masses m and carrying charges q1 and q2 are connected by a cord of length l and can slide without friction on the wires. Considering the case when the beads are stationary, determine.
(a) the angle α = ∠APQ,
(b) the tension in the cord
(c) the normal reactions on the beads.
(d) If the cord is now cut, what are the values of the charges for which the beads continue to remains stationary?

Sol.:
1. (a) Forces acting on each bead: mg downwards, tension T along the cord, electric force F = \( \frac{K q_1 q_2}{l^2} \) along the line joining the beads, and normal reactions from the wires.
2. Resolving along and perpendicular to the wires for equilibrium of the beads gives: T cos α = F cos α + mg sin 30° ....(i)
3. F sin α + N1 = mg cos 30° + T sin α ....(ii)
T sin α = F sin α + mg sin 60° ....(iii)
N2 + F cos α = T cos α + mg cos 60° ....(iv)
4. From (i) and (iii): (T - F) cos α = mg sin 30° ....(v) and (T - F) sin α = mg cos 30° ....(vi)
5. Dividing (v) by (vi): cot α = tan 30° = cot 60°, so α = 60°.
6. (b) Substituting α = 60° in (i): T cos 60° = \( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{l^2} \cos 60° + mg \sin 30° \)
7. T = \( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{l^2} + mg \)
8. (c) From (ii): N1 = mg cos 30° + (T - F) sin 60° = mg \( \frac{\sqrt{3}}{2} \) + mg \( \frac{\sqrt{3}}{2} \) = √3 mg
9. From (iv): N2 = (T - F) cos 60° + mg cos 60° = mg \( \frac{1}{2} \) + mg \( \frac{1}{2} \) = mg
10. (d) When the cord is cut, T = 0. From (i): 0 = F cos 60° + mg sin 30°, or F + mg = 0. This result is on the assumption that q1 and q2 are of the same sign, so a force of repulsion was taken. Since mg is fixed in direction, its sign cannot be reversed, but the sign of F can be reversed: if q1 and q2 are of opposite sign, F changes its sign from + to -.
11. Let q1 and q2 be of opposite sign; then -F + mg = 0 is the condition for equilibrium, that is, \( \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{l^2} = mg \), so |q1q2| = 4πε0 mgl².
12. Hence q1q2 = 4πε0 mgl², with the charges of opposite sign.

Teacher's Note:
a) The two beads lie on the two inclined wires, so their weights make angles of 30° and 60° with the wires.
b) The cord tension equals the electric force plus mg when the cord is taut.
c) With the cord cut, the electric force alone must balance the weight, so the charges must attract.

Ex.14 A thin conducting ring of radius R has a linear charge density λ = λ0 cos φ, where λ0 is the maximum linear density and φ is angle measured anticlockwise from OA. Calculate the electric field at the centre of the ring.

Sol.:
1. Consider an element EF of length R dφ at angle φ. Its charge is dq = (λ0 cos φ)(R dφ).
2. Field at O due to this element: dE = \( \frac{dq}{4\pi\epsilon_0 R^2} = \frac{\lambda_0 R \cos\phi \, d\phi}{4\pi\epsilon_0 R^2} = \frac{\lambda_0 \cos\phi \, d\phi}{4\pi\epsilon_0 R} \)
3. Resolving into components: dE_x = -\( \frac{\lambda_0 \cos^2\phi \, d\phi}{4\pi\epsilon_0 R} \) and dE_y = -\( \frac{\lambda_0 \sin\phi \cos\phi \, d\phi}{4\pi\epsilon_0 R} \)
4. Integrating over 0 to 2π: E_x = -\( \frac{\lambda_0}{4\pi\epsilon_0 R} \int_0^{2\pi} \cos^2\phi \, d\phi = -\frac{\lambda_0}{4\pi\epsilon_0 R} \times \pi \)
5. E_y = -\( \frac{\lambda_0}{4\pi\epsilon_0 R} \int_0^{2\pi} \sin\phi \cos\phi \, d\phi = 0 \)
6. So E = -\( \frac{\lambda_0}{4\epsilon_0 R} \hat{i} \), and its magnitude is |E| = \( \frac{\lambda_0}{4\epsilon_0 R} \).

Teacher's Note:
a) The y-components cancel because the charge distribution is symmetric about the x-axis.
b) The integral of cos²φ over a full circle equals π, which gives the factor of one-quarter.
c) The field points from the positive half of the ring towards the negative half, along the negative x-direction.

Question. The current produced in wire when 10⁷ electron/sec are flowing in it:
a. 1.6 × 10^–26 amp
b. 1.6 × 10¹² amp
c. 1.6 × 10²⁶ amp
d. 1.6 × 10^–12 amp
Answer : D

Question. The force between two charges 0.06 m apart is 5 N . If each charge is moved towards the other by 0.01m , then the force between them will become:
a. 7.20 N
b. 11.25 N
c. 22.50 N
d. 45.00 N
Answer : B

Question. The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of 0.036 N/C just above its surface is:
a. 2.7 ×10⁵
b. 2.6 ×10⁵
c. 2.5 ×10⁵
d. 2.4 ×10⁵
Answer : C

Question. Three particles, each having a charge of 10 μC are placed at the corners of an equilateral triangle of side 10 cm . The electrostatic potential energy of the system is: (Given 1/4πε₀ = 9 x 109 N - m² / C²)
a. Zero
b. Infinite
c. 27 J
d. 100 J
Answer : C

Question. Two equal spheres are identically charged with q units of electricity separately. When they are placed at a distance 3R from centre-to-centre where R is the radius of either sphere the force of repulsion between them is:

Answer : A

Question. The insulation property of air breaks down at E = 3×10⁶ volt/metre. The maximum charge that can be given to a sphere of diameter 5 m, is approximately: (in coulombs)
a. 2 x 10⁻²
b. 2 x 10⁻³
c. 2 x 10⁻⁴
d. 2 x 10⁻⁵
Answer : B

Question. q, 2q, 3q and 4q charges are placed at the four corners A, B, C and D of a square. The field at the centre O of the square has the direction along:

a. AB
b. CB
c. AC
d. BD
Answer : B

Question. What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude newton/coulomb 2 [1/ 4πε₀ = 9×10⁹ Nm²]
a. 11 2 x 10⁻¹¹ coulomb
b. 11 3 x 10⁻¹¹ coulomb
c. 11 5 x 10⁻¹¹ coulomb
d. 11 9 x 10⁻¹¹ coulomb
Answer : A

Question. Two point charges Q and –3Q are placed at some distance apart. If the electric field at the location of Q is E, then at the locality of – 3Q, it is:
a. –E
b. E/3
c. –3E
d. - E/3
Answer : B

Question. Two charged spheres of radius R₁ and R₂ respectively are charged and joined by a wire. The ratio of electric field of the spheres is:
a. R₁/R₂
b. R₂/R₁
c. R₁²/R₂²
d. R₂²/R₁
Answer : B

Question. Equal charges Q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point A is:
a. Q/4πε₀a²
b. √2Q/4πε₀a²
c. √3Q/4πε₀a²
d. Q/2πε₀a²
Answer : C

Question. Eight charges having the valves as shown are arranged symmetrically on a circle of radius 0.4m in air. Potential at centre O will be:

a. 63×10⁴ volt
b. 10 63×10¹⁰ volt
c. 63×10⁶ volt
d. Zero
Answer : A

Question. As shown in the figure, charges +q and –q are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is:

Answer : D

Question. A hollow metal sphere of radius 5cm is charged such that the potential on its surface is 10 volts. The potential at the centre of the sphere is:
a. Zero
b. 10 V
c. Same as at a point 5 cm away from the surface
d. Same as at a point 25 cm away from the surface
Answer : B

Question. A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is:

a. 54 e.s.u.
b. 1/4 e.s.u.
c. 30 e.s.u.
d. 36 e.s.u.
Answer : D

Question. A charge Q is distributed over two concentric hollow spheres of radii r and (R > r) such that the surface densities are equal. The potential at the common centre is:
a. Q(R² + r²) / 4πε₀(R + r)
b.  Q/R + r
c. Zero
d. Q(R + r) / 4πε₀(R² + r²)
Answer : D

Question. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of – 3Q, the new potential difference between the two surfaces is:
a. V
b. 2V
c. 4V
d. –2V
Answer : A

Question. An electric dipole of moment p placed in a uniform electric field E has minimum potential energy when the angle between p and E is:
a. Zero
b. π/2
c. π
d. 3π/2
Answer : C

Question. If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x : y is:
a. 1 : 1
b. 1 : 2
c. 1 : 2
d. 2 : 1
Answer : D

Question. Three charges of (+2q), (– q) and (– q) are placed at the corners A, B and C of an equilateral triangle of side a as shown in the adjoining figure. Then the dipole moment of this combination is:

a. qa
b. Zero
c. q a√3
d. (2/√3) qa
Answer : C

Sol.: When these four charges are lying at infinity initially, these can be brought from infinity to the respective points A, B, C and D. Thus work has to be done in assembling these
charges because of their interaction with each other. The total work done will be equal to the sum of potential energies required for each pair of particles.

The following forces act on the sphere :
(i) Weight = mg
(ii) Electrostatic force of repulsion between sphere and metallic plate = F
(iii) Tension in the string = T
The string is deflected through angle θ to the vertical. When the sphere is in equilibrium state, resultant force