JEE Physics Electrostatic Potential and Capacitance MCQs Set 12

Answer: (d) 20 N/C

Explanation:
1. Moving the charge slowly, the external work equals the gain in potential energy: \( W=qE\,d\cos60^\circ \).
2. 4 = 0.2 × E × 2 × ½ = 0.2E, so E = 20 N/C.

Teacher's Note:
1. Only the displacement component along the field counts.
2. Slow motion means no change in kinetic energy.

Answer: (b) A, B and C

Explanation:
1. At A: the two nearest charges (+q and −q) are both at a/2, and the other two (+q and −q) are both at a√5/2, so their potentials cancel in pairs: V = 0. The same holds at C.
2. At B all four charges are equidistant and the algebraic sum is zero.
3. At P the two +q charges are nearer than the two −q charges, so V > 0 (and V < 0 at Q).
4. Hence V = 0 at A, B and C.

Teacher's Note:
1. Potential is a scalar: add the contributions with their signs.
2. Pair equal distances to find the cancellations.

Answer: (a) 27 V

Explanation:
1. Every element of the ring is at distance R from the centre.
2. \( V=\frac{kQ}{R}=\frac{9\times10^9\times1.5\times10^{-9}}{0.5}=27\ \text{V} \).

Teacher's Note:
1. Potential at the centre depends only on total charge and radius.
2. The shape (semicircle or full circle) does not matter.

Answer: (a) 8 V

Explanation:
1. E = F/q = 2000/5 = 400 V/m.
2. V = E d = 400 × 0.02 = 8 V.

Teacher's Note:
1. Field = force per unit charge.
2. Use V = Ed for a uniform field.

Answer: (c) Eb

Explanation:
1. The potential difference depends only on the displacement component parallel to the field.
2. The horizontal displacement a is perpendicular to E and contributes nothing.
3. So |V_A − V_B| = Eb.

Teacher's Note:
1. Resolve the displacement along and across the field.
2. Equipotentials are perpendicular to the field lines.

Answer: (c) \( \sqrt{2mQV} \)

Explanation:
1. QV = ½mv², so v = √(2QV/m).
2. Momentum p = mv = √(2mQV).

Teacher's Note:
1. Use the work–energy theorem.
2. Multiply the speed by m.

Answer: (a) V/6r

Explanation:
1. Potential difference: \( \frac{kq}{r}-\frac{kq}{3r}=\frac{2kq}{3r}=V \), so kq = 3rV/2.
2. Field at 3r: \( E=\frac{kq}{9r^2}=\frac{3rV/2}{9r^2}=\frac{V}{6r} \).

Teacher's Note:
1. Find kq from the potential difference.
2. Then use E = kq/(distance)².

Answer: (c) may increase or decrease

Explanation:
1. Potential energy U = qV.
2. For a positive charge it decreases; for a negative charge it increases.
3. So it may do either, depending on the sign of the charge.

Teacher's Note:
1. The sign of the charge decides.
2. U = qV.

Answer: (c) may increase or decrease

Explanation:
1. U = kq₁q₂/r.
2. For like charges, U rises as r decreases; for unlike charges, U falls (becomes more negative).
3. So it may increase or decrease.

Teacher's Note:
1. The sign of q₁q₂ matters.
2. Compare like and unlike charges.

Answer: (a) −2/3

Explanation:
1. With equal separations d: \( U=\frac{k}{d}\left[q\cdot2q+q\cdot xq+2q\cdot xq\right]=\frac{kq^2}{d}(2+3x)=0 \).
2. So x = −2/3.

Teacher's Note:
1. Add the three pair energies.
2. Each pair is at the same distance d.

Answer: (d) n^(5/3) U

Explanation:
1. Self-energy of a sphere is \( U\propto\frac{q^2}{R} \).
2. The new drop has charge nq and radius n^(1/3)R (volume conservation).
3. So \( U'=\frac{n^2}{n^{1/3}}U=n^{5/3}U \).

Teacher's Note:
1. Charges add and volumes add.
2. Use U ∝ q²/R.

Answer: (d) −500 N

Explanation:
1. Near the origin the slope is +2.5 V/cm = 250 V/m, so E = −250 V/m.
2. F = qE = 2 × (−250) = −500 N.

Teacher's Note:
1. Convert V/cm to V/m.
2. Force = q × E, with the sign of E.

Answer: (b) V_x = −xE₀

Explanation:
1. \( V=-\int_0^xE_0\,dx=-E_0x \).

Teacher's Note:
1. Potential falls along the direction of the field.
2. V = −∫E·dx.

Answer: (b) is proportional to r²

Explanation:
1. Let E = kr, directed outward.
2. \( V(r)=-\int_0^rkr\,dr=-\frac{kr^2}{2} \).
3. So the potential is proportional to r² (and decreases outward).

Teacher's Note:
1. Integrate E with the correct sign.
2. Linear field gives quadratic potential.

Answer: (b) 1 : 2

Explanation:
1. qV = ½mv², so v ∝ √q.
2. v_A : v_B = √q : √(4q) = 1 : 2.

Teacher's Note:
1. Use energy gained = qV.
2. Speed varies as the square root of the charge.

Answer: (d) r/4

Explanation:
1. At closest approach all kinetic energy is converted: \( \tfrac12mv^2=\frac{kQq}{r} \).
2. Doubling v multiplies the left side by 4, so r becomes r/4.

Teacher's Note:
1. Energy conservation from infinity to the closest point.
2. r ∝ 1/v².

Answer: (a) V

Explanation:
1. The potential difference between the sphere and the shell depends only on the charge Q on the inner sphere: \( kQ\left(\frac1a-\frac1b\right) \) with b the inner radius of the shell, plus the equal potential of the shell material.
2. Adding charge to the shell shifts the potential of the shell and of everything inside it by the same amount.
3. So the potential difference is unchanged and equals V.

Teacher's Note:
1. A charge on the outer shell raises all potentials inside by a constant.
2. Differences of potential are unaffected.

Answer: (b) \( \sqrt{\frac{2eV}{m}} \)

Explanation:
1. eV = ½mv².
2. So v = √(2eV/m).

Teacher's Note:
1. Work done by the field equals kinetic energy gained.
2. Solve for v.

Answer: (d) \( \frac{Q}{4\pi\varepsilon_0R} \)

Explanation:
1. Every surface element is at the same distance R from the centre.
2. \( V=\int\frac{k\,dq}{R}=\frac{kQ}{R}=\frac{Q}{4\pi\varepsilon_0R} \), whatever the distribution.

Teacher's Note:
1. Potential is a scalar sum of dq/R contributions.
2. Only the total charge Q enters.

Answer: (a) √3 u

Explanation:
1. Potential varies as ln r: between ℓ and 2ℓ the change in KE is \( \tfrac12m(2u^2-u^2)=\tfrac12mu^2=C\ln2 \) for a constant C.
2. From ℓ to 4ℓ the change is C ln 4 = 2C ln 2 = mu².
3. So \( \tfrac12mv^2=\tfrac12mu^2+mu^2 \), giving v² = 3u² and v = √3 u.

Teacher's Note:
1. For a line charge, equal ratios of distance give equal energy changes.
2. Doubling the distance ratio doubles the energy change.

Answer: (c) positive

Explanation:
1. B stays neutral overall, but the induced negative charge is nearer to A than the induced positive charge.
2. The potential at B is dominated by the positive charge on A and is shifted upward: the net potential of B is positive.
3. It is the same throughout B because B is a conductor.

Teacher's Note:
1. Induced charges sum to zero, so only the external charge A changes the potential of B.
2. A positive charge raises the potential of nearby conductors.