JEE Physics Electrostatic Potential and Capacitance MCQs Set 06

Answer: (c) 2 times

Explanation:
1. Volume is conserved: \( \frac43\pi R^3=8\cdot\frac43\pi r^3\Rightarrow R=2r \).
2. For a sphere \( C=4\pi\varepsilon_0R\propto R \), so the big drop has 2 times the capacitance.

Teacher's Note:
1. Capacitance of an isolated sphere is proportional to its radius.
2. Volume (not area) is conserved when drops merge.

Answer: (b) 4 : 1

Explanation:
1. Big drop radius \( R=2r \) and charge \( Q=8q \).
2. \( \frac{V_{big}}{V_{small}}=\frac{kQ/R}{kq/r}=\frac{8}{2}=4 \).

Teacher's Note:
1. Potential of a sphere is kQ/R.
2. Charge adds, radius scales as the cube root of the number of drops.

Answer: (d) \( V\times N^{2/3} \)

Explanation:
1. Big radius \( R=N^{1/3}r \), total charge \( Q=Nq \) with \( q=4\pi\varepsilon_0rV \).
2. \( V'=\frac{kQ}{R}=\frac{Nkq}{N^{1/3}r}=N^{2/3}V \).

Teacher's Note:
1. Charge grows as N, radius as N^{1/3}.
2. The potential therefore grows as N^{2/3}.

Answer: (d) 2 μF, 18 μF

Explanation:
1. Minimum: all three in series, \( \frac{6}{3}=2 \) μF.
2. Maximum: all three in parallel, \( 3\times6=18 \) μF.

Teacher's Note:
1. Series gives the minimum, parallel the maximum.
2. For n equal capacitors the extremes are C/n and nC.

Answer: (b) CV/2

Explanation:
1. Total charge CV is conserved and shared equally between identical capacitors.
2. Each carries \( \frac{CV}{2} \) at a common potential \( \frac V2 \).

Teacher's Note:
1. Charge is conserved when isolated capacitors are joined.
2. Half the stored energy is lost as heat and radiation in the connecting process.

Answer: (b) 2222 V

Explanation:
1. Series: \( C_{eq}=2 \) pF, so each capacitor carries \( Q=2\times5000=10^4 \) pC.
2. Total charge when joined in parallel: \( 2Q=2\times10^4 \) pC; capacitance \( 3+6=9 \) pF.
3. \( V=\frac{2\times10^4}{9}=2222 \) V.

Teacher's Note:
1. Capacitors in series carry equal charges.
2. In the parallel reconnection the charges of the joined plates add.

Answer: (c) 200 μC

Explanation:
1. \( \frac1C=\frac13+\frac1{10}+\frac1{15}=\frac{10+3+2}{30}=\frac12 \), so \( C=2 \) μF.
2. \( Q=CV=2\times100=200 \) μC, the same on each capacitor in series.

Teacher's Note:
1. In a series circuit the charge is common to all capacitors.
2. The charge equals C_eq × V.

Answer: (b) \( 1500\sqrt2 \) volt

Explanation:
1. Distance of each corner from the centre \( r=\frac{8\sqrt2}{2}=4\sqrt2 \) cm \( =0.0566 \) m.
2. \( V=4\times\frac{9\times10^9\times(10/3)\times10^{-9}}{4\sqrt2\times10^{-2}}=\frac{120}{4\sqrt2\times10^{-2}}=1500\sqrt2 \) V.

Teacher's Note:
1. Potential is a scalar, so the four contributions simply add.
2. Half of the diagonal is the distance from the centre to a corner.

Answer: (c) \( \frac{\sqrt2Q^2}{\pi\varepsilon_0a} \)

Explanation:
1. Potential at centre due to the four charges: \( V=4\cdot\frac{kQ}{a/\sqrt2}=\frac{4\sqrt2kQ}{a} \).
2. Initial potential energy of −Q: \( U_i=-QV \); final \( U_f=0 \).
3. \( W=U_f-U_i=QV=\frac{4\sqrt2kQ^2}{a}=\frac{\sqrt2Q^2}{\pi\varepsilon_0a} \).

Teacher's Note:
1. Work done by the external agent equals the increase in potential energy.
2. Here the removed charge is attracted by the others, so positive work has to be done.

Answer: (d) \( -\frac{9}{4} \) J

Explanation:
1. AC = 50 cm (3-4-5 triangle). The 100 μC charge at A is fixed.
2. \( W=q[V_C-V_B]=5\times10^{-6}\times9\times10^9\times100\times10^{-6}\left(\frac{1}{0.5}-\frac{1}{0.4}\right) \)
3. \( =4.5\times(2-2.5)=-\frac94 \) J.

Teacher's Note:
1. Work done only depends on the initial and final distances from the fixed charge.
2. C is farther from A (50 cm) than B (40 cm), so the repulsion itself does positive work and the external agent does negative work.

Answer: (a) −q/2

Explanation:
1. \( U=k\left[\frac{q\cdot q}{l}+\frac{Qq}{l}+\frac{Qq}{l}\right]=\frac{k}{l}(q^2+2Qq)=0 \).
2. \( Q=-\frac{q}{2} \).

Teacher's Note:
1. Energy of a system is the sum over every pair, counted once.
2. A zero total needs Q to be opposite in sign to q.

Answer: (a) 8 along negative X-axis

Explanation:
1. \( E_x=-\frac{\partial V}{\partial x}=-8x \), and \( E_y=E_z=0 \).
2. At x = 1 m: \( E=-8 \) V/m, i.e. 8 V/m along the negative X-axis.

Teacher's Note:
1. The field is the negative gradient of the potential.
2. Ignore the coordinates that V does not depend on.

Answer: (d) None of these

Explanation:
1. \( W=qEd\cos60° \).
2. \( 4=0.2\times E\times2\times0.5=0.2E\Rightarrow E=20 \) N/C, which is not among options (a) to (c).

Teacher's Note:
1. Only the component of displacement along the field does work.
2. If your answer is not among the options, recheck the substitution once and then pick 'none of these'.

Answer: (a) 10 V

Explanation:
1. \( E=\frac{F}{q}=\frac{5000}{5}=1000 \) N/C.
2. \( \Delta V=Ed=1000\times0.01=10 \) V.

Teacher's Note:
1. Find the field first, then multiply by the separation along the field line.
2. Always convert cm to m.

Answer: (b) 1 : 2

Explanation:
1. \( qV=\frac12mv^2\Rightarrow v=\sqrt{\frac{2qV}{m}} \).
2. \( \frac{v_A}{v_B}=\sqrt{\frac{q}{4q}}=\frac12 \).

Teacher's Note:
1. Speed gained scales as the square root of the charge for equal mass and voltage.
2. Don't forget the square root.

Answer: (d) r/4

Explanation:
1. At the closest point all kinetic energy becomes potential energy: \( \frac12mv^2=\frac{kqQ}{r} \).
2. Doubling the speed quadruples the kinetic energy, so \( r'=\frac{r}{4} \).

Teacher's Note:
1. Closest approach: KE converts entirely to potential energy.
2. r varies inversely with the initial kinetic energy.

Answer: (b) A, B and C

Explanation:
1. The line through A, B and C is the perpendicular bisector of each +q/−q pair, so each point on it is equidistant from +q and −q partners and the potential is zero there.
2. P is nearer the two +q charges than the −q charges, so \( V_P>0 \); similarly \( V_Q<0 \).

Teacher's Note:
1. Points equidistant from equal and opposite charges are at zero potential.
2. P and Q are on the axis through the charges, not on the bisector.

Answer: (b) 90 m/s

Explanation:
1. Energy conservation: \( \frac12mv^2=kq_1q_2\left(\frac11-\frac1{10}\right) \).
2. \( kq_1q_2=9\times10^9\times10^{-6}\times10^{-3}=9 \), so \( \frac12mv^2=9\times0.9=8.1 \) J.
3. \( v=\sqrt{\frac{2\times8.1}{2\times10^{-3}}}=90 \) m/s.

Teacher's Note:
1. Convert the mass to kg (2 g = 2 × 10⁻³ kg).
2. Potential energy lost by the pair appears as kinetic energy.

Answer: (b) r (n − 1)

Explanation:
1. With all n corners filled the field at the centre is zero, so the field of the missing corner's charge is cancelled: the actual field is the field of one charge, \( E=\frac{kq}{r^2} \).
2. Potential from (n−1) charges: \( V=\frac{(n-1)kq}{r} \).
3. \( \frac VE=(n-1)r \).

Teacher's Note:
1. Use the symmetry trick: full set gives zero field, so (n−1) charges equal minus the missing charge.
2. Potential has no such cancellation, it is a plain sum.

Answer: (d) \( -8\hat i+4\hat j \)

Explanation:
1. The equipotential line has direction \( \hat i+2\hat j \) (slope 2).
2. The field is perpendicular to it: \( \vec E\cdot(\hat i+2\hat j)=0 \).
3. Only \( -8\hat i+4\hat j \) gives \( -8+8=0 \).

Teacher's Note:
1. E is always perpendicular to the equipotential surface.
2. Test each option with a dot product.

Answer: (a) −20 volt/m

Explanation:
1. \( E=-\frac{dV}{dx}=-(10x+10) \).
2. At x = 1 m: \( E=-20 \) V/m.

Teacher's Note:
1. Differentiate first, substitute afterwards.
2. Don't forget the minus sign in E = −dV/dx.

Answer: (b) 0.88 mm

Explanation:
1. \( E=\frac{\sigma}{2\varepsilon_0}=\frac{10^{-7}}{2\times8.85\times10^{-12}}=5.65\times10^3 \) V/m.
2. \( d=\frac{\Delta V}{E}=\frac{5}{5650}=8.8\times10^{-4} \) m = 0.88 mm.

Teacher's Note:
1. A non-conducting sheet gives σ/2ε₀ on each side.
2. Equipotential surfaces near a sheet are parallel planes.

Answer: (d) \( \left(\frac{R_1}{R_2}\right)^2 \)

Explanation:
1. For a charged conducting sphere \( V=ER \) (since \( E=\frac{kQ}{R^2} \) and \( V=\frac{kQ}{R} \)).
2. \( \frac{V_1}{V_2}=\frac{E_1R_1}{E_2R_2}=\frac{R_1}{R_2}\cdot\frac{R_1}{R_2}=\left(\frac{R_1}{R_2}\right)^2 \).

Teacher's Note:
1. On the surface of a sphere V = E R.
2. Substitute the given field ratio directly.

Answer: (b) −q/4

Explanation:
1. Let the inner sphere carry charge x. The uncharged shell has zero net charge, so it contributes zero potential at the centre.
2. Earthed inner sphere: \( V=\frac{kx}{a}+\frac{kq}{4a}=0 \).
3. \( x=-\frac q4 \).

Teacher's Note:
1. Potential at the centre is a sum of contributions: the sphere's own charge, the shell and the outside point charge.
2. A shell with zero net charge adds nothing to the potential inside it.