QJEEM15Ph5ec4: JEE Main 2015 Physics - Class 12 JEEM15Ph5ec4 Solution: A uniformly charged solid sphere of radius has potential…

Question

Verified Solution • 2025–26 Session moderate

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 potentials

{{3{V_0}} \over 2},\,{{5{V_0}} \over 4},\,{{3{V_0}} \over 4}

and

{{{V_0}} \over 4}

have radius

{R_1},\,\,{R_2},\,\,{R_3}

and

{R_4}

respectively. Then

A ${R_1} = 0$ and ${R_2} < \left( {{R_4} - {R_3}} \right)$
B $2R < {R_4}$
C ${R_1} = 0$ and ${R_2} > \left( {{R_4} - {R_3}} \right)$
D ${R_1} \ne 0$ and $\left( {{R_2} - {R_1}} \right) > \left( {{R_4} - {R_3}} \right)$

Accepted Answer

(a,b)

We know,

{V_0} = {{Kq} \over R} = Vsurface

Now,

{V_i} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {R^2}} \right)\,\,\,\,\,

[For

r &lt; R

]

At the center of sphere

r=0.

Here

V = {3 \over 2}{V_0}

Now,

{5 \over 4}{{Kq} \over R} = {{Kq} \over {2{R^3}}}\left( {3{R^2} - {r^2}} \right)

{R_2} = {R \over {\sqrt 2 }}

{3 \over 4}{{Kq} \over R} = {{Kq} \over {{R^3}}}

{1 \over 4}{{Kq} \over R} = {{Kq} \over {R{}_4}}

{R_4} = 4R

Also,

{R_1} = 0

and

{R_2} &lt; \left( {{R_4} - R{}_3} \right)

Step-by-Step Verified Solution