Question

The equipotential surfaces corresponding to a single positive charge are concentric spherical shells with the charge at its origin. The spacing between the surfaces for the same change in potential:

• A. is uniform throughout the field
• B. is getting closer as $r\to \infty$
• C. is getting closer as $r\to 0$
• D. can be varied as one wishes to

Answer 1

Answer: (C)

is getting closer as $r\to 0$

Let the equipotential surfaces have potential $V_1,V_2,V_{3}with,V_1>V_2>V_3$
Also, let $V_3-V_2=V_2-V_1=R$
Potential of a charge is given by $kQ/r$
$V=kQ/r\Rightarrow r=kQ/V$

Distance between the equipotential surfaces is

$kQ(\frac{1}{V_2}-\frac{1}{V_3})$ and $kQ(\frac{1}{V_1}-\frac{1}{V_2})$
or
$kQ\left(\frac{V_3-V_2}{V_3{V}_2}\right)$ and $kQ\left(\frac{V_2-V_1}{V_2{V}_1}\right)$
or
$kQ\left(\frac{R}{V_3{V}_2}\right)$ and $kQ\left(\frac{R}{V_2{V}_1}\right)$
Clearly $V_2{V}_1>V_3{V}_2$
Thus
$kQ\left(\frac{R}{V_3{V}_2}\right)>kQ\left(\frac{R}{V_2{V}_1}\right)$
Therefore spacing between surfaces decreases as potential increases i.e. $r$ decreases.