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Electrostatics Part 1

Section: Physics

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Question : 65 of 100

Marks: +1, -0

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

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

\frac{V_0}{4}

R_1,R_2,R_3

R_4

$R_1\neq 0$ and $(R_2-R_1)>(R_4-R_3)$
$R_1=0$ and $R_2<(R_4-R_3)$
$2R < R_4$
$R_1=0$ and $R_2>(R_4-R_3)$

Solution:

The potential at the centre

=\frac{kQ}{\frac{4}{3}\pi R^3} \int\limits_0^R \frac{4\pi r^3 dr}{r} = \frac{3}{2} \frac{kQ}{R} = \frac{3}{2} V_0; k = \frac{1}{4\pi\varepsilon_0}

\therefore R_1=0

Potential at surface,

V_0=\frac{KQ}{R}

Potential at

R_2=\frac{5V_0}{4} \Rightarrow R_2=\frac{R}{\sqrt{2}}

Potential at

R_3,\frac{kQ}{R_3}=\frac{3}{4}\frac{KQ}{R} \Rightarrow R_3=\frac{4R}{3}

Similarly at

R_4,\frac{kQ}{R_4}=\frac{kQ}{4R} \Rightarrow R_4=4R

∴ Both options (2) and (3) are correct.

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