JEE Main · 2015 · 04 Apr 2015 · Offline · Electrostatics

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A uniformly charged solid sphere of radius R has potential V0 (measured with respect to infinity) on its surface. For this sphere the equipotential surfaces with potentials 3V0/2, 5V0/4, 3V0/4 and V0/4 have radius R1, R2, R3 and R4 respectively. Then: [options — see source PDF]

(A) $R_{1} \neq = 0$ and $(R_{2} - R_{1}) > (R_{4} - R_{3})$
(B) $R_{1} = 0$ and $R_{2} < (R_{4} - R_{3})$ ✓ Correct
(C) $2 R > R_{4}$
(D) $R_{1} = 0$ and $R_{2} > (R_{4} - R_{3})$

Answer: (B)

Solution

For a uniformly charged solid sphere: inside (

r \leq R

)

V = \frac{V _{0}}{2} (3 - \frac{r ^{2}}{R ^{2}})

(so

V = 3 V_{0} /2

at the centre,

V_{0}

at the surface); outside (

r \geq R

)

V = V_{0} R / r

. Then

3 V_{0} /2

is the maximum potential, reached only at the centre, so

R_{1} = 0

. For

5 V_{0} /4

(inside):

\frac{V _{0}}{2} (3 - r^{2} / R^{2}) = \frac{5 V _{0}}{4} \Rightarrow r = R / 2

, so

R_{2} = R / 2

. For

3 V_{0} /4

(outside):

V_{0} R / r = 3 V_{0} /4 \Rightarrow R_{3} = 4 R /3

. For

V_{0} /4

(outside):

V_{0} R / r = V_{0} /4 \Rightarrow R_{4} = 4 R

. Thus

R_{4} - R_{3} = 8 R /3 \approx 2.67 R

, which exceeds

R_{2} = R / 2 \approx 0.71 R

. Hence

R_{1} = 0

and

R_{2} < (R_{4} - R_{3})

— option (B).

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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JEE Main · 2015 · 04 Apr 2015 · Offline · Electrostatics

Solution

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