CBSE Class 12 Physics Electrostatics Frequently Asked Questions

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FREQUENTLY ASKED QUESTIONS
CLASS 12 PHYSICS : ELECTROSTATICS

1 MARKS

1) Force of attraction between two point charges placed at a distance of 'd' is 'F'. What distance apart they are kept in the same medium, so that, the force between them is 'F/3'? [1 Mark]

Answer: The distance must be √3 d, since F is proportional to 1/r² and tripling the distance gives F/3 when r becomes √3 d.

Teacher's Note:
a) Force follows the inverse square law, so the distance changes by the square root of the force ratio.

2) Define electric field intensity. Write its S I unit. Write the magnitude and direction of electric field intensity due electric dipole of length 2a at the midpoint of the line joining the two charges. [1 Mark]

Answer:
1. Electric field intensity at a point is the force experienced per unit positive test charge placed at that point, E = F/q₀.
2. SI unit: newton per coulomb (N C^-1), equivalent to volt per metre (V m^-1).
3. Magnitude at the midpoint: E = 2kq/a², where k = 1/(4πε₀) and q is the magnitude of each charge.
4. Direction: along the line joining the charges, from the positive charge towards the negative charge (parallel to the dipole axis from +q to -q).

Teacher's Note:
a) Both charges give fields in the same direction at the midpoint, so the magnitudes add, not cancel.

3) Define electric field intensity. Write its S.I unit. Write the magnitude and direction of electric field intensity due to an electric dipole of length2a at the midpoint of the line joining the two charges. [1 Mark]

Answer:
1. Electric field intensity at a point is the force per unit positive test charge placed at that point, E = F/q₀.
2. SI unit: newton per coulomb (N C^-1).
3. Magnitude at the midpoint of the dipole: E = 2kq/a², where k = 1/(4πε₀).
4. Direction: from the positive charge towards the negative charge, along the dipole line.

Teacher's Note:
a) The two fields point the same way at the midpoint, so they add directly.

4) Sketch the electric lines of force due to point charges q > 0, q < 0 and for uniform field [1 Mark]

Answer:

Field lines for a positive point charge point radially outward, for a negative point charge point radially inward, and for a uniform field they are parallel, equally spaced straight lines.

Teacher's Note:
a) Arrows show the direction of the force on a positive test charge, and lines never cross.

5) Define electric flux. Give its S.I unit and dimensional formula. [1 Mark]

Answer: Electric flux through a surface is the total number of electric field lines crossing that surface normally, given by φ = ∮E·dS. Its SI unit is N m² C^-1 and its dimensional formula is [M L³ T^-3 A^-1].

Teacher's Note:
a) Flux is a scalar quantity, obtained from the dot product of E and dS.

6) Two point charges 4μc and -2μc are separated by a distance of 1 m in air. At what point on the line joining the charges is the electric potential zero? [1 Mark]

Answer:
1. Let the point be at a distance x from the 4 μC charge, between the charges, so its distance from the -2 μC charge is (1 - x) m.
2. Potential zero: k(4 × 10^-6)/x + k(-2 × 10^-6)/(1 - x) = 0.
3. 4/x = 2/(1 - x), so 4(1 - x) = 2x.
4. 4 - 4x = 2x, so x = 2/3 m.
5. Hence the potential is zero at 2/3 m (about 0.67 m) from the 4 μC charge, that is, 1/3 m from the -2 μC charge.

Teacher's Note:
a) Potential is a scalar, so add the two potentials with their signs.

7) Depict the equipotential surfaces for a system of two identical positive point charges placed at distance d apart. [1 Mark]

Answer:

Around each positive charge the equipotential surfaces are closed surfaces, roughly spherical near that charge. Farther away, the surfaces surround both charges together. At the midpoint, the equipotential surface has a saddle-shaped section where the potential is the same as the midpoint value.

Teacher's Note:
a) Equipotential surfaces are always perpendicular to the electric field lines.

8) Deduce the expression for the potential energy of a system of two point charges q1 and q2 brought from infinity to that points r1 and r2. [1 Mark]

Answer:
1. Work to bring q1 from infinity to r1 is zero, since no field exists yet.
2. Potential at r2 due to q1 is V = kq1/r12, where r12 is the distance between r1 and r2.
3. Work to bring q2 from infinity to r2 is q2 × kq1/r12.
4. Potential energy U = kq1q2/r12 = (1/(4πε₀)) × q1q2/|r1 - r2|.

Teacher's Note:
a) The potential energy of the system is the total work done in assembling the charges, and it is stored in the system.

2 MARKS

9) Derive an expression for electric field intensity at a point on the axial line and on the equatorial line of an electric pole. [2 Marks]

Answer:
1. Let the dipole have charges +q and -q separated by 2a, and dipole moment p = 2aq, with the point P at distance r from the centre.
2. Axial point: E_axial = kq/(r - a)² - kq/(r + a)² = \( \frac{4kqar}{(r^2 - a^2)^2} \).
3. Hence E_axial = \( \frac{2kpr}{(r^2 - a^2)^2} \), which for r much greater than a becomes \( \frac{2kp}{r^3} \), along the dipole moment.
4. Equatorial point: each charge gives a field kq/(r² + a²), and their components along the axis add while the perpendicular components cancel.
5. E_equatorial = \( \frac{kp}{(r^2 + a^2)^{\frac{3}{2}}} \), which for r much greater than a becomes \( \frac{kp}{r^3} \), opposite to the dipole moment.

Teacher's Note:
a) Use the distance from each charge, not from the centre, when writing the field of each charge.
b) The axial field is twice the equatorial field for r much greater than a.

10) Derive an expression for torque acting on an electric dipole in a uniform electric filed. [2 Marks]

Answer:
1. Forces on the two charges are +qE and -qE, and they are equal and opposite.
2. The perpendicular distance between the lines of action of the two forces is 2a sin θ.
3. Torque τ = qE × 2a sin θ.
4. Since p = 2aq, τ = pE sin θ, or in vector form τ = p × E.

Teacher's Note:
a) The net force on the dipole in a uniform field is zero, so only torque acts.
b) Use θ as the angle between p and E, not the angle with the axis.

11) Derive an expression for total work done in rotating an electric dipole through an angle 'θ' in uniform electric field. [2 Marks]

Answer:
1. Torque at angle θ is τ = pE sin θ.
2. Small work done dW = τ dθ = pE sin θ dθ.
3. Total work W = ∫₀^θ pE sin θ dθ = pE(1 - cos θ).
4. Hence the work done is W = pE(1 - cos θ), assuming the dipole starts from the position parallel to the field.

Teacher's Note:
a) Work done is stored as potential energy U = -pE cos θ.
b) If the starting angle is not zero, use W = pE(cos θ1 - cos θ2).

12) A sphere 'S1' of radius 'r1' encloses a charge 'Q'. If there is another concentric sphere S2 of the radius r2 (r2 > r1) and there be no additional charges between S1 and S2, find the ratio of electric flux through S1 and S2. [2 Marks]

Answer:
1. By Gauss's law, the flux through a closed surface depends only on the charge enclosed by it.
2. Flux through S1 = Q/ε₀, since it encloses Q.
3. Flux through S2 = Q/ε₀, since it also encloses the same charge Q only.
4. Ratio of flux through S1 to S2 = 1 : 1.

Teacher's Note:
a) The radius of the surface does not change the flux, only the enclosed charge does.
b) Students often think the bigger sphere has less flux because the field is weaker there.

13) State Gauss's Theorem in electrostatics. Using this theorem, find the electric field strength due to an infinite plane sheet of charge. [2 Marks]

Answer:
1. Gauss's theorem: the total electric flux through a closed surface equals the net charge enclosed divided by ε₀.
2. Take a cylindrical Gaussian surface with its axis normal to the sheet, with cross-section area A.
3. Flux through the two flat ends = 2EA, and no flux through the curved side.
4. 2EA = σA/ε₀, so E = \( \frac{\sigma}{2\varepsilon_0} \).

Teacher's Note:
a) The field is uniform on both sides of an infinite sheet and points normally away from it.
b) The charge enclosed is σA, only the part of the sheet inside the cylinder.

14) State Gauss' theorem. Apply this theorem to obtain the expression for the electric field intensity at a point due to an infinitely long, thin, uniformly charged straight wire. [2 Marks]

Answer:
1. Gauss's theorem: the total electric flux through any closed surface is equal to 1/ε₀ times the net charge enclosed by it.
2. Take a cylinder of radius r and length l, coaxial with the wire, with linear charge density λ.
3. Charge enclosed = λl, and flux through the curved surface = E × 2πrl.
4. E × 2πrl = λl/ε₀, so E = \( \frac{\lambda}{2\pi\varepsilon_0 r} \), directed radially outward for positive λ.

Teacher's Note:
a) No flux passes through the two flat ends, since the field is radial and parallel to them.
b) The field falls off as 1/r, not as 1/r², for a line charge.

15) Using Gauss's theorem, show mathematically that for any point outside the shell, the field due to a uniformly charged thin spherical shell is the same as if the entire charge of the shell is concentrated at the centre. Why do you expect the electric field inside the shell to be zero according to this theorem? [2 Marks]

Answer:
1. Let the shell have radius R and total charge Q. Take a concentric spherical Gaussian surface of radius r greater than R.
2. By symmetry, E is the same at all points of this surface and is directed radially.
3. Flux = E × 4πr² = Q/ε₀, so E = \( \frac{Q}{4\pi\varepsilon_0 r^2} \).
4. This is the same as the field of a point charge Q at the centre, so the shell acts as if its charge is at the centre.
5. Inside the shell (r less than R), the Gaussian sphere encloses no charge, so the flux is zero and E = 0.

Teacher's Note:
a) Charge on a conductor sits on its outer surface, so no charge lies inside the shell.
b) The symmetry of the shell makes the field direction radial everywhere.

16) Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric of a dipole at a point as compared to that due to single charge. [2 Marks]

Answer:
1. Let P be on the axis at distance r from the centre of the dipole.
2. Potential at P: V = kq/(r - a) - kq/(r + a).
3. V = \( \frac{2kqa}{r^2 - a^2} \) = \( \frac{kp}{r^2 - a^2} \), and for r much greater than a, V ≈ \( \frac{kp}{r^2} \).
4. Contrast: the potential of a dipole falls off as 1/r², whereas the potential of a single charge falls off as 1/r.

Teacher's Note:
a) Potential is found by adding the scalar potentials of the two charges, with their signs.
b) The faster fall-off of dipole potential is a common exam point.

17) Define dielectric constant in terms of the capacitance of a capacitor. [2 Marks]

Answer:
1. Dielectric constant K is the ratio of the capacitance with a dielectric filling the space to the capacitance of the same capacitor with air or vacuum between the plates.
2. K = C/C₀, where C is the capacitance with the dielectric and C₀ is the capacitance in air or vacuum.

Teacher's Note:
a) K is a pure number with no unit.
b) The geometry of the capacitor must be the same in both cases.

3 MARKS

18) Briefly explain the principle of a capacitor. Derive an expression for the capacitance of a parallel plate capacitor, whose plates are separated by a dielectric medium. [3 Marks]

Answer:
1. Principle: two conductors placed close together, carrying equal and opposite charges, store energy in the field between them. Bringing the plates close lowers the potential difference, so more charge can be stored.
2. Let the plates have area A, separation d, charge density σ, and the dielectric constant be K. Field in dielectric E = σ/(Kε₀).
3. Potential difference V = Ed = \( \frac{\sigma d}{K\varepsilon_0} \), and charge Q = σA.
4. C = Q/V = \( \frac{\sigma A}{\sigma d/(K\varepsilon_0)} \) = \( \frac{K\varepsilon_0 A}{d} \).

Teacher's Note:
a) A capacitor stores charge and energy, and its capacitance depends only on its geometry and the medium.
b) Use the field inside the dielectric, which is smaller than the field in air by a factor K.
c) The result reduces to ε₀A/d when K = 1.

19) Derive an expression for the energy stored in a parallel plate capacitor with air between the plates. How does the stored energy change if air is replaced by a medium of dielectric constant 'K'? ; Also show that the energy density of a capacitor is. [3 Marks]

Answer:
1. Work done to charge the capacitor to a charge q at potential V is dW = (q/C) dq.
2. Total energy U = ∫₀^Q (q/C) dq = Q²/(2C) = ½CV².
3. For air, C = ε₀A/d, so U = \( \frac{1}{2}\frac{\varepsilon_0 A}{d}V^2 \).
4. If the charge Q is held constant and the air is replaced by a dielectric of constant K, C becomes KC, so U' = Q²/(2KC) = U/K. The stored energy falls by a factor K.
5. Energy density u = U/(Ad) = \( \frac{1}{2}\varepsilon_0 E^2 \), since V = Ed and U = ½ (ε₀A/d)(Ed)².

Teacher's Note:
a) The question does not state whether the battery stays connected, so the charge is taken as constant.
b) If the battery stays connected, V stays constant and the energy becomes K times larger instead.
c) Energy density depends only on the field, not on the area or separation.

20) A parallel-plate capacitor is charged to a potential difference V by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change
(i) electric field between the plates
(ii) capacitance, and
(iii) rgy stored in the capacitor [3 Marks]

Answer:
1. The capacitor is disconnected, so the charge Q stays constant.
2. (i) Electric field E = σ/ε₀ = Q/(ε₀A) depends only on charge and area, so it remains unchanged.
3. (ii) Capacitance C = ε₀A/d halves when d is doubled.
4. (iii) Energy U = Q²/(2C) doubles, since C is halved.

Teacher's Note:
a) With the source removed, charge is the quantity that stays fixed.
b) The field does not depend on the separation when the charge is fixed.
c) Energy increases because work must be done against the attraction of the plates while they are pulled apart.

21) Explain the underlying principle of working of a parallel plate capacitor. If two similar plates, each of area 'A' having surface charge densities '+ σ' & '- σ' are separated by a distance ' d' in air, write expressions for (i) the electric field at points between the two plates, (ii) the potential difference between the plates & (iii) the capacity of the capacitor so formed [3 Marks]

Answer:
1. Principle: when two conducting plates carry equal and opposite charges, the field is confined between them, and the potential difference between the plates allows the system to store energy.
2. (i) Electric field between the plates: E = σ/ε₀, directed from the positive plate to the negative plate.
3. (ii) Potential difference: V = Ed = \( \frac{\sigma d}{\varepsilon_0} \).
4. (iii) Capacitance: C = Q/V = \( \frac{\sigma A}{\sigma d/\varepsilon_0} \) = \( \frac{\varepsilon_0 A}{d} \).

Teacher's Note:
a) The field outside the two plates is zero, since the two equal and opposite fields cancel there.
b) Capacity here means capacitance, measured in farad.
c) Keep the area and separation in the final expression for the capacity.

22) A parallel plate capacitor is charged by a battery and the battery remains connected, a dielectric slab is inserted in the space between the plates. Explain what changes if any , occur in the values of
(I) potential difference between the plates
(II) electric field between the plates
(III) nergy stored in the capacitor. [3 Marks]

Answer:
1. The battery remains connected, so the potential difference V stays fixed.
2. (I) Potential difference between the plates remains the same, equal to V.
3. (II) Electric field E = V/d remains unchanged, since V and d are both fixed.
4. (III) Capacitance becomes KC, so the charge rises to KQ and the energy U = ½CV² becomes K times the original value, that is, it increases.

Teacher's Note:
a) The battery supplies extra charge, which is why the stored charge increases.
b) The field inside the dielectric is smaller, but the battery keeps the field equal to V/d.
c) Energy increases here, unlike the case where the capacitor is isolated.