JEE Physics Electrostatic Potential and Capacitance MCQs Set 20

Answer: (b) 2 : 1

Explanation:
1. After connection both are at the same potential V.
2. The surface field is \( E=\dfrac{V}{r} \).
3. \( E_A/E_B=r_B/r_A=2/1 \).

Teacher's Note:
1. Smaller sphere, larger field: the principle of the lightning conductor.
2. Charge density also varies inversely with radius.

Answer: (d) rE

Explanation:
1. \( E=\dfrac{kq}{(r/2)^2}=\dfrac{4kq}{r^2} \).
2. Potential due to both: \( V=2\cdot\dfrac{kq}{r/2}=\dfrac{4kq}{r}=rE \).

Teacher's Note:
1. Potential falls as 1 over distance, field as 1 over distance squared.
2. Both spheres contribute equally to the potential.

Answer: (a) \( r_1/r_2 \)

Explanation:
1. For a conducting sphere \( V=Er \) at the surface, and the centre has the same potential.
2. So \( V_1/V_2=r_1/r_2 \) when \( E_1=E_2 \).

Teacher's Note:
1. Potential inside a conductor equals its surface potential.
2. Surface field is surface potential over radius.

Answer: (d) \( \dfrac{1}{4\pi\epsilon_0}\dfrac{Q-q}{b} \)

Explanation:
1. Induced charge: −Q on the inner surface and −q + Q on the outer surface.
2. Within the metal the potential is constant and equals the surface potential, determined by the net charge Q − q seen from outside: \( V=\dfrac{k(Q-q)}{b} \).

Teacher's Note:
1. A conductor is an equipotential.
2. Evaluate the potential at the outer surface, where only the total charge matters.

Answer: (c) −Q on the inner surface and Q − q on the outer surface

Explanation:
1. The field inside the metal must vanish, so the inner surface carries −Q to cancel the central charge's flux.
2. The total on the shell is −q, so the outer surface carries \( -q-(-Q)=Q-q \).

Teacher's Note:
1. Gauss's law on a surface inside the metal fixes the inner charge.
2. The remainder goes to the outer surface.

Answer: (c) R/3

Explanation:
1. Potential at the centre of a uniformly charged solid sphere is \( \dfrac{3kQ}{2R} \).
2. Half of it is \( \dfrac{3kQ}{4R} \). Outside, \( \dfrac{kQ}{r}=\dfrac{3kQ}{4R} \) so \( r=4R/3 \).
3. Distance from the surface \( =\dfrac{4R}3-R=\dfrac R3 \).

Teacher's Note:
1. Inside the sphere, the potential never falls to half, so the point is outside.
2. Do not forget to subtract R.

Answer: (a) \( \dfrac{\sigma}{\epsilon_0}(b-a-c) \)

Explanation:
1. Charges: \( Q_A=-4\pi a^2\sigma,\ Q_B=4\pi b^2\sigma,\ Q_C=-4\pi c^2\sigma \).
2. \( V_A=k\left[\dfrac{Q_A}{a}+\dfrac{Q_B}{b}+\dfrac{Q_C}{c}\right]=\dfrac{\sigma}{\epsilon_0}(-a+b-c) \).

Teacher's Note:
1. At A, each shell contributes kQ over its own radius.
2. Surface density times 4π radius squared gives charge.

Answer: (c) \( 0,\ 0,\ Q+q_1+q_2 \)

Explanation:
1. On connection all conductors reach the same potential.
2. Charge placed on a conductor inside a closed hollow conductor flows to its outer surface, so A and B retain no charge.
3. All the charge \( Q+q_1+q_2 \) ends up on C.

Teacher's Note:
1. Inside a cavity the field vanishes once all charge is on the outer surface.
2. Connected conductors form one conductor.

Answer: (b) \( \dfrac{Kq}{6a} \)

Explanation:
1. Let C hold charge x. Earthing: \( V_C=K\left[\dfrac{q}{3a}+\dfrac{x}{3a}-\dfrac{q}{4a}\right]=0 \) gives \( x=-q/4 \).
2. A is uncharged and lies inside all shells: \( V_A=K\left[\dfrac{q}{2a}+\dfrac{x}{3a}-\dfrac{q}{4a}\right]=\dfrac{Kq}{a}\left[\dfrac12-\dfrac1{12}-\dfrac14\right]=\dfrac{Kq}{6a} \).
3. Since \( V_C=0 \), \( V_A-V_C=\dfrac{Kq}{6a} \).

Teacher's Note:
1. Outside a shell the potential is that of a point charge, inside it is constant.
2. Find the charge on the earthed shell from the condition V = 0.

Answer: (c) \( \dfrac{\lambda q_0}{\pi\epsilon_0}\ln2 \)

Explanation:
1. For a line charge \( V=-\dfrac{\lambda}{2\pi\epsilon_0}\ln r \).
2. \( V_A=\dfrac{\lambda}{2\pi\epsilon_0}\ln\dfrac{2a}{a}=\dfrac{\lambda\ln2}{2\pi\epsilon_0} \) and \( V_B=\dfrac{\lambda}{2\pi\epsilon_0}\ln\dfrac{a}{2a}=-\dfrac{\lambda\ln2}{2\pi\epsilon_0} \).
3. \( W=q_0(V_A-V_B)=\dfrac{\lambda q_0\ln2}{\pi\epsilon_0} \).

Teacher's Note:
1. Superpose the potentials of the two wires; the constants cancel in the difference.
2. Work by the field is charge times the fall in potential.

Answer: (c) \( \dfrac{1}{4\pi\epsilon_0}\dfrac{3Q}{r^2} \)

Explanation:
1. Apply Gauss's law to a sphere of radius r between the two: the enclosed charge is only the +3Q of the inner sphere.
2. \( E=\dfrac{1}{4\pi\epsilon_0}\dfrac{3Q}{r^2} \).

Teacher's Note:
1. The outer shell does not contribute to the field inside it.
2. Spherical symmetry gives the inverse-square form.

Answer: (d) 2.5 kV/m

Explanation:
1. Outside, \( E\propto1/r^2 \).
2. The diameter is 2R, so the field is \( 10\times\left(\dfrac{R}{2R}\right)^2=2.5 \) kV/m.

Teacher's Note:
1. Compare distances from the centre.
2. Keep units in kV/m.

Answer: (a) \( n^{-2/3}V \)

Explanation:
1. Charge per drop \( q/n \); volume is conserved so radius \( r'=r/n^{1/3} \).
2. \( V'=\dfrac{kq/n}{r/n^{1/3}}=n^{-2/3}V \).

Teacher's Note:
1. Volume is conserved when a drop divides.
2. Potential is charge divided by radius.

Answer: (c) 4 times

Explanation:
1. Radius of the large drop is 2r and its charge is 8q.
2. \( V'=\dfrac{k\,8q}{2r}=4\cdot\dfrac{kq}{r} \).

Teacher's Note:
1. Volume conservation gives radius 2r.
2. Charge adds up directly.

Answer: (d) \( \sqrt{\dfrac{\rho R^2}{4m\epsilon_0}} \)

Explanation:
1. Inside, \( V(r)=\dfrac{\rho(3R^2-r^2)}{6\epsilon_0} \).
2. The ends are at r = R; the middle of the chord is at r = R/2, where V is highest.
3. \( \Delta V=\dfrac{\rho}{6\epsilon_0}\left[\left(3R^2-\dfrac{R^2}{4}\right)-2R^2\right]=\dfrac{\rho R^2}{8\epsilon_0} \).
4. \( \tfrac12mv^2=1\cdot\Delta V \) gives \( v=\sqrt{\dfrac{\rho R^2}{4m\epsilon_0}} \).

Teacher's Note:
1. The particle only has to climb to the highest-potential point on the path.
2. Potential inside a uniformly charged sphere decreases with r.

Answer: (a) (ii) and (iii) only

Explanation:
1. Placing A inside B induces −Q on the inner surface and +Q on the outer surface of B, which changes the potential of A, so (i) is false.
2. Earthing B drains the +Q on the outer surface, so (ii) is true. A's potential then depends only on the pair A and the inner surface of B, which is lower than before, so (iii) is true and (iv) false.

Teacher's Note:
1. Earthed outer surface sheds its charge.
2. Removing +Q from the outer surface lowers the potential of everything inside.

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. \( \dfrac{kQ}{R+0.05}=100 \) and \( \dfrac{kQ}{R+0.10}=75 \) give \( R=0.10\ \text{m} \) and \( kQ=15\ \text{V m} \).
2. Surface potential \( =15/0.10=150\ \text{V} \) (i); \( Q=15/(9\times10^9)=\tfrac53\times10^{-9}\ \text{C} \), so (ii) is false; \( E=15/0.01=1500\ \text{V/m} \) (iii) true; centre potential \( \tfrac32\times150=225\ \text{V} \) (iv) true.

Teacher's Note:
1. Two equations in R and kQ.
2. Centre potential of a uniformly charged solid sphere is 1.5 times the surface value.

Answer: (a) E/4

Explanation:
1. Surface density \( \sigma=\dfrac{Q}{4\pi R^2} \) becomes one-fourth when R doubles.
2. At the surface of a conductor \( E=\sigma/\epsilon_0 \), so the field becomes E/4.

Teacher's Note:
1. Surface field is proportional to the local surface charge density.
2. Area grows as the square of the radius.

Answer: (b) \( -Q\ \text{and}\ -\dfrac{qR}{l} \)

Explanation:
1. The inner surface must carry −Q so that the field in the metal is zero (Gauss's law).
2. The earthed sphere keeps zero potential, which with the outside charge q at distance l fixes the charge on the outer surface at \( -\dfrac{qR}{l} \) (image-charge result), the cavity charge Q being completely shielded.

Teacher's Note:
1. The cavity charge's effect is confined to the inner surface plus its compensating +Q on the outer surface, which the earth removes.
2. Earthing sets the sphere's potential to zero; the potential due to q at the sphere is kq/l.

Answer: (d) \( \dfrac Q2-\epsilon_0EA\ \text{ and }\ \dfrac Q2+\epsilon_0EA \)

Explanation:
1. The field is cancelled inside the metal, so each face carries a charge of \( \pm\epsilon_0EA \) induced by the external field (negative where the field lines end, positive where they begin).
2. The original charge Q shares equally between the two faces.

Teacher's Note:
1. Induced charge density on a conductor is ε₀ times the normal field.
2. Add the induced and the shared original charge.

Answer: (b) \( 7.5\times10^{-9}\ \text{N} \)

Explanation:
1. Speed: \( v=\sqrt{\dfrac{2eV}{m}}=\sqrt{\dfrac{2\times1.6\times10^{-19}\times500}{9\times10^{-31}}}=1.33\times10^7\ \text{m/s} \).
2. Electrons per second \( n=\dfrac{I}{e}=6.25\times10^{14} \).
3. Force = rate of change of momentum \( =nmv=6.25\times10^{14}\times9\times10^{-31}\times1.33\times10^7=7.5\times10^{-9}\ \text{N} \).

Teacher's Note:
1. Force equals the momentum delivered to the surface per second.
2. The electrons stop completely, so the whole momentum mv is transferred.

Answer: (a) 3

Explanation:
1. \( E=V/d=\dfrac{1500}{1.5\times10^{-2}}=10^5\ \text{V/m} \).
2. Balance \( neE=mg \): \( n=\dfrac{4.9\times10^{-15}\times9.8}{1.6\times10^{-19}\times10^5}=3 \).

Teacher's Note:
1. The upward electric force on negative charges balances the weight.
2. The result must be a whole number.

Answer: (d) \( 9V_0 \)

Explanation:
1. Big-drop radius \( R=3r \) (volume conservation) and charge \( Q=27q \).
2. \( V=\dfrac{k\cdot27q}{3r}=9\cdot\dfrac{kq}{r}=9V_0 \).

Teacher's Note:
1. Volume is conserved, so radius grows by the cube root of 27.
2. Potential is charge over radius.

Answer: (b) \( -\dfrac{qr_1}{r_2} \)

Explanation:
1. Let the inner shell carry x after earthing; its potential must be zero: \( \dfrac{kx}{r_1}+\dfrac{kq}{r_2}=0 \).
2. So \( x=-\dfrac{qr_1}{r_2} \).

Teacher's Note:
1. The outer shell's total charge q stays constant; only its distribution changes.
2. Potential inside a shell is constant and equal to that on the shell.

Answer: (d) \( \dfrac{q}{4\pi\epsilon_0r} \)

Explanation:
1. The outer surface carries +q uniformly distributed (the cavity charge is shielded and the outer surface is spherical).
2. Outside, the potential equals that of a point charge q at the centre: \( V=\dfrac{kq}{r} \).

Teacher's Note:
1. The outer surface charge distributes uniformly regardless of where q sits in the cavity.
2. The potential from the inner surface (−q) and the cavity charge (+q) cancel outside.