CBSE Class 12 Physics Electric Charges and Fields MCQs Set 05

Answer: (a) 5 × 10¹³

Explanation:
1. Using q = ne, n = \( \frac{q}{e} \) = \( \frac{8 \times 10^{-6}}{1.6 \times 10^{-19}} \)
2. n = 5 × 10¹³ electrons

Teacher's Note:
1. A negative charge means extra electrons; a positive charge means electrons are missing.
2. Convert μC to C (× 10⁻⁶) before dividing by e.

Answer: (c) 2.4 × 10⁻¹⁹ C

Explanation:
1. Charge must be an integral multiple of e = 1.6 × 10⁻¹⁹ C.
2. 3.2 × 10⁻¹⁹ = 2e, 4.8 × 10⁻¹⁹ = 3e and 1.6 × 10⁻¹⁸ = 10e.
3. 2.4 × 10⁻¹⁹ C = 1.5e, which is not a whole number of e. So it is not possible.

Teacher's Note:
1. Divide each value by 1.6 × 10⁻¹⁹; only whole-number results are allowed.
2. This is a direct test of quantisation of charge.

Answer: (c) 16F

Explanation:
1. F ∝ \( \frac{q_1 q_2}{r^2} \)
2. New force = \( \frac{(2q_1)(2q_2)}{(r/2)^2} \) ∝ \( \frac{4 q_1 q_2}{r^2/4} \) = 16 × \( \frac{q_1 q_2}{r^2} \)
3. New force = 16F

Teacher's Note:
1. Doubling both charges multiplies F by 4; halving r multiplies it by another 4.
2. Write the ratio F′/F instead of putting in numbers.

Answer: (b) 5 N

Explanation:
1. F_medium = \( \frac{F_{\text{air}}}{K} \)
2. F_medium = \( \frac{20}{4} \) = 5 N

Teacher's Note:
1. A medium always reduces the Coulomb force, so 80 N is a trap.
2. The direction (repulsion) does not change in the medium.

Answer: (a) 0.6 N

Explanation:
1. F = k \( \frac{q_1 q_2}{r^2} \) = \( \frac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.3)^2} \)
2. F = \( \frac{54 \times 10^{-3}}{0.09} \) = 0.6 N (repulsive)

Teacher's Note:
1. Always convert cm to m: 30 cm = 0.3 m.
2. Both charges are positive, so the force is repulsive.

Answer: (b) 900 N C⁻¹

Explanation:
1. E = k \( \frac{q}{r^2} \) = \( \frac{9 \times 10^9 \times 4 \times 10^{-9}}{(0.2)^2} \)
2. E = \( \frac{36}{0.04} \) = 900 N C⁻¹

Teacher's Note:
1. 1 nC = 10⁻⁹ C, which cancels neatly with k = 9 × 10⁹.
2. The field points away from the charge because it is positive.

Answer: (c) They form closed loops.

Explanation:
1. Electrostatic field lines begin on positive charges and end on negative charges, so they cannot close on themselves.
2. Statement (c) is therefore not correct; the other three are true properties.

Teacher's Note:
1. Closed loops occur for magnetic field lines, not electrostatic ones.
2. Lines cannot cross because the field has one direction at each point.

Answer: (a) 8 × 10⁻¹² C m, from −2 nC to +2 nC

Explanation:
1. p = q × 2a = 2 × 10⁻⁹ × 4 × 10⁻³ = 8 × 10⁻¹² C m
2. The dipole moment points from the negative charge to the positive charge.
3. Answer: 8 × 10⁻¹² C m, from −2 nC to +2 nC

Teacher's Note:
1. 2a is the full separation between the charges, not half of it.
2. Remember the direction: p points from −q to +q.

Answer: (b) 2 : 1

Explanation:
1. E_axial = k \( \frac{2p}{r^3} \) and E_equatorial = k \( \frac{p}{r^3} \)
2. Ratio = 2 : 1

Teacher's Note:
1. Both fields fall as 1/r³; only the factor 2 differs.
2. The axial field is along p, the equatorial field is opposite to p.

Answer: (a) 1 × 10⁻⁴ N m

Explanation:
1. τ = pE sin θ = 4 × 10⁻⁹ × 5 × 10⁴ × sin 30°
2. τ = 2 × 10⁻⁴ × 0.5 = 1 × 10⁻⁴ N m

Teacher's Note:
1. Use sin θ, not cos θ, for torque.
2. 2 × 10⁻⁴ N m is the maximum torque, reached only at 90°.

Answer: (b) −226 N m² C⁻¹

Explanation:
1. Only enclosed charges count: q_enc = 3 − 5 = −2 nC
2. φ = \( \frac{q_{\text{enc}}}{\varepsilon_0} \) = \( \frac{-2 \times 10^{-9}}{8.854 \times 10^{-12}} \) ≈ −226 N m² C⁻¹

Teacher's Note:
1. The +4 nC charge outside contributes zero net flux.
2. A negative flux means more field lines enter the surface than leave it.

Answer: (c) r⁰ (independent of r)

Explanation:
1. By Gauss's law, E = \( \frac{\sigma}{2\varepsilon_0} \)
2. There is no r in the expression, so E is independent of the distance.

Teacher's Note:
1. The field lines of an infinite sheet are parallel, so the field does not weaken with distance.
2. Compare: line charge ∝ 1/r, point charge ∝ 1/r², dipole ∝ 1/r³.

Answer: (d) 4 × 10⁵ N C⁻¹

Explanation:
1. E = \( \frac{\lambda}{2\pi\varepsilon_0 r} \) = \( \frac{2k\lambda}{r} \)
2. E = \( \frac{2 \times 9 \times 10^9 \times 2 \times 10^{-6}}{0.09} \) = 4 × 10⁵ N C⁻¹

Teacher's Note:
1. Rewrite 1/(2πε₀) as 2k to make the arithmetic easy.
2. The field is directed radially outward for a positive line charge.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Charge is quantised: q = ne, so the Assertion is correct.
2. Charge is transferred by whole electrons, so the charge can only change in steps of e. The Reason is correct.
3. The Reason explains why charge is quantised, so it is the correct explanation.

Teacher's Note:
1. The value of e is 1.6 × 10⁻¹⁹ C.
2. At the macroscopic level quantisation can be ignored because n is huge.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. φ = EA, so its unit is (N C⁻¹)(m²) = N m² C⁻¹. The Assertion is correct.
2. Field lines do start from positive charges, so the Reason is correct.
3. The direction in which field lines start has nothing to do with the unit of flux, so the Reason does not explain the Assertion.

Teacher's Note:
1. The unit of flux can also be written V m.
2. When the Reason is true but about a different idea, choose (b).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. If two lines crossed, there would be two tangents, i.e. two field directions, at the crossing point.
2. The field at a point is a single vector with one direction, so lines cannot cross.
3. Both statements are correct and the Reason explains the Assertion.

Teacher's Note:
1. The tangent to a field line gives the direction of E.
2. This property holds for any number of source charges.

Answer: (c) A is true, but R is false.

Explanation:
1. The forces qE and −qE are equal and opposite, so the net force is zero. The Assertion is correct.
2. The two forces act at different points and form a couple of torque τ = pE sin θ, which is not zero unless θ = 0° or 180°. The Reason is false.

Teacher's Note:
1. Zero net force does not mean zero torque.
2. A dipole in a non-uniform field feels both a net force and a torque.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For a Gaussian sphere of radius r < R, q_enc = 0, so φ = E × 4πr² = 0.
2. By symmetry E has the same magnitude everywhere on that sphere, so E = 0.
3. Both statements are correct and the Reason explains the Assertion.

Teacher's Note:
1. All the charge of a shell lies on its surface.
2. This is why the field inside a hollow charged conductor is zero.

Answer: (d) A is false, but R is true.

Explanation:
1. F_water = \( \frac{F_{\text{air}}}{K} \) = \( \frac{F_{\text{air}}}{80} \), so the force becomes 80 times smaller. The Assertion is false.
2. The dielectric constant of water is about 80, so the Reason is true.

Teacher's Note:
1. A dielectric medium always weakens the force between charges.
2. This is why water dissolves ionic salts so easily.

Answer: (a) 3.6 N

Explanation:
1. F = k \( \frac{q_1 q_2}{r^2} \) = \( \frac{9 \times 10^9 \times 4 \times 10^{-6} \times 4 \times 10^{-6}}{(0.2)^2} \)
2. F = \( \frac{0.144}{0.04} \) = 3.6 N (attractive)

Teacher's Note:
1. Use the magnitudes of the charges; the signs only tell you the force is attractive.
2. 20 cm = 0.2 m.

Answer: (c) 7.2 × 10⁶ N C⁻¹, towards B

Explanation:
1. Distance of O from each sphere = 0.1 m. Field due to each = \( \frac{9 \times 10^9 \times 4 \times 10^{-6}}{(0.1)^2} \) = 3.6 × 10⁶ N C⁻¹
2. The field of +4 μC at O points away from A (towards B); the field of −4 μC points towards B.
3. Both are in the same direction, so E = 2 × 3.6 × 10⁶ = 7.2 × 10⁶ N C⁻¹, towards B

Teacher's Note:
1. Between unlike charges the two fields add; between like charges they subtract.
2. 'Zero' would be correct only if both charges were equal and of the same sign.

Answer: (d) 7.2 N, towards B

Explanation:
1. F = qE = 1 × 10⁻⁶ × 7.2 × 10⁶ = 7.2 N
2. A positive charge moves along the field, so the force is 7.2 N, towards B.

Teacher's Note:
1. Once you know E, the force on any charge is simply qE.
2. It is repelled by A and attracted by B, which agrees with the answer.

Answer: (b) 1.8 N

Explanation:
1. F_oil = \( \frac{F_{\text{air}}}{K} \) = \( \frac{3.6}{2} \) = 1.8 N

Teacher's Note:
1. The medium divides the force by K.
2. The charges and their separation are unchanged.

Answer: (a) 8 × 10⁻⁷ C m, directed from B to A

Explanation:
1. p = q × 2a = 4 × 10⁻⁶ × 0.2 = 8 × 10⁻⁷ C m
2. p points from the negative charge (B) to the positive charge (A).
3. Answer: 8 × 10⁻⁷ C m, from B to A

Teacher's Note:
1. The dipole moment points opposite to the field at the midpoint.
2. Use the full separation 0.2 m as 2a.

Answer: (c) 226 N m² C⁻¹

Explanation:
1. The sphere of radius 20 cm encloses the whole charge, 2 nC.
2. φ = \( \frac{2 \times 10^{-9}}{8.854 \times 10^{-12}} \) ≈ 226 N m² C⁻¹

Teacher's Note:
1. Flux depends only on the enclosed charge, not on the radius of the Gaussian surface.
2. Dividing by 8.854 × 10⁻¹² is the same as multiplying by about 1.13 × 10¹¹.

Answer: (a) zero

Explanation:
1. A Gaussian sphere of radius 5 cm lies inside the shell and encloses no charge.
2. So the flux through it is zero and, by symmetry, E = 0.

Teacher's Note:
1. 7200 N C⁻¹ is what you would get by wrongly using the point-charge formula at r = 5 cm.
2. The field is zero everywhere inside a charged shell.

Answer: (b) 200 N C⁻¹

Explanation:
1. Outside the shell, E = k \( \frac{q}{r^2} \) = \( \frac{9 \times 10^9 \times 2 \times 10^{-9}}{(0.3)^2} \)
2. E = \( \frac{18}{0.09} \) = 200 N C⁻¹

Teacher's Note:
1. Outside, the shell behaves like a point charge at its centre.
2. Measure r from the centre, not from the surface.

Answer: (d) remains the same

Explanation:
1. Both spheres enclose the same charge, 2 nC.
2. φ = q_enc/ε₀ does not depend on the size of the surface, so the flux remains the same.

Teacher's Note:
1. The field weakens as 1/r², but the area grows as r², so E × area stays constant.
2. This is the idea behind Gauss's law.

Answer: (c) 1800 N C⁻¹

Explanation:
1. Just outside the surface, r = R = 0.1 m.
2. E = \( \frac{9 \times 10^9 \times 2 \times 10^{-9}}{(0.1)^2} \) = 1800 N C⁻¹

Teacher's Note:
1. The field is greatest just outside the surface and drops to zero just inside it.
2. This value also equals σ/ε₀.