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A uniformly charged solid sphere of radius R has potential V 0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3 V 0 2 , 5 V 0 4 , 3 V 0 4 and V 0 4 have radius R 1 , R 2 , R 3 and R 4 respectively. Then Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

Option D: R 1 = 0 and R 2 R v = 1 4 π ε 0 Q R on the surface v = 1 4 π ε 0 Q R 3 2 - 1 2 r 2 R 2 inside i.e. r < R Clearly 3 v 0 2 , 5 v 0 4 are inside potentials ∵ > v 0 3 v 0 4 , v 0 4 are outside potentials ∵ < v 0 To get R 1 : 3 v 0 2 = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2 where, v 0 = 1 4 π ε 0 Q R 3 2 × 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 1 2 R 2 3 2 = 3 2 - 1 2 R 1 2 R 2 ⇒ R 1 = 0 To get R 2 : 5 4 v 0 = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2 5 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 2 - 1 2 R 2 2 R 2 5 4 = 3 2 - 1 2 R 2 2 R 2 1 2 R 2 2 R 2 = 1 4 ⇒ R 2 = R 2 To get R 3 : 3 v 0 4 = 1 4 π ε 0 Q R 3 3 4 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 3 3 4 R = 1 R 3 R 3 = 4 3 R To get R 4 : v 0 4 = 1 4 π ε 0 Q R 4 1 4 × 1 4 π ε 0 Q R = 1 4 π ε 0 Q R 4 R 4 = 4 R R 4 - R 3 = 4 R - 4 R 3 = 8 R 3 > R 2 R 1 = 0 and R 2 < R 4 - R 3

JEE Main 2015PhysicsElectrostaticsMediumMCQ

JEE Main 2015 — Electrostatics Question with Solution

JEE Main 2015 (04 Apr)

A uniformly charged solid sphere of radius R has potential $V_{0}$ (measured with respect to $\infty$ ) on its surface. For this sphere the equipotential surfaces with potential $\frac{3 V_{0}}{2}, \frac{5 V_{0}}{4}, \frac{3 V_{0}}{4}$ and $\frac{V_{0}}{4}$ have radius $R_{1}$ , $R_{2}, R_{3}$ and $R_{4}$ respectively. Then

Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it.

Show full solutionCorrect option: D

Correct answer

R_{1} = 0

and

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

Step-by-step explanation

Potential for uniformly charged solid sphere

v = \frac{1}{4 π ε_{0}} \frac{Q}{r}

outside i.e r > R

v = \frac{1}{4 π ε_{0}} \frac{Q}{R}

on the surface

v = \frac{1}{4 π ε_{0}} \frac{Q}{R} [\frac{3}{2} - \frac{1}{2} \frac{r^{2}}{R^{2}}]

inside i.e. r < R

Clearly

\frac{3 v_{0}}{2}, \frac{5 v_{0}}{4}

are inside potentials

[∵ > v_{0}]

\frac{3 v_{0}}{4}, \frac{v_{0}}{4}

are outside potentials

[∵ < v_{0}]

To get

R_{1}

\frac{3 v_{0}}{2} = \frac{1}{4 π ε_{0}} \frac{Q}{R} [\frac{3}{2} - \frac{1}{2} \frac{R_{1}^{2}}{R^{2}}]

{where, v}_{0} = \frac{1}{4 π ε_{0}} \frac{Q}{R}

\frac{3}{2 \times 4 π ε_{0}} \frac{Q}{R} = \frac{1}{4 π ε_{0}} \frac{Q}{R} [\frac{3}{2} - \frac{1}{2} \frac{R_{1}^{2}}{R^{2}}]

\frac{3}{2} = \frac{3}{2} - \frac{1}{2} \frac{R_{1}^{2}}{R^{2}} \Rightarrow R_{1} = 0

To get

R_{2}

\frac{5}{4} v_{0} = \frac{1}{4 π ε_{0}} \frac{Q}{R} [\frac{3}{2} - \frac{1}{2} \frac{R_{2}^{2}}{R^{2}}]

\frac{5}{4} \frac{1}{4 π ε_{0}} \frac{Q}{R} = \frac{1}{4 π ε_{0}} \frac{Q}{R} [\frac{3}{2} - \frac{1}{2} \frac{R_{2}^{2}}{R^{2}}]

\frac{5}{4} = \frac{3}{2} - \frac{1}{2} \frac{R_{2}^{2}}{R^{2}}

\frac{1}{2} \frac{R_{2}^{2}}{R^{2}} = \frac{1}{4}

\Rightarrow

R_{2} = \frac{R}{\sqrt{2}}

To get

R_{3}

\frac{3 v_{0}}{4} = \frac{1}{4 π ε_{0}} \frac{Q}{R_{3}}

\frac{3}{4} \frac{1}{4 π ε_{0}} \frac{Q}{R} = \frac{1}{4 π ε_{0}} \frac{Q}{R_{3}}

\frac{3}{4 R} = \frac{1}{R_{3}}

R_{3} = \frac{4}{3} R

To get

R_{4}

\frac{v_{0}}{4} = \frac{1}{4 π ε_{0}} \frac{Q}{R_{4}}

\frac{1}{4} \times \frac{1}{4 π ε_{0}} \frac{Q}{R} = \frac{1}{4 π ε_{0}} \frac{Q}{R_{4}}

R_{4} = 4 R

R_{4} - R_{3} = 4 R - \frac{4 R}{3} = \frac{8 R}{3} > R_{2}

R_{1} = 0

and

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

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This is a previous-year question from JEE Main 2015, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.

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