Discussion Forum

A uniformly charged solid sphere of radius R has potential V₀ (measured with respect to ∞) on its surface. For this sphere, the equipotential sufaces,  with potentials  $\frac{3V_{0}}{2},\frac{5V_{0}}{4},\frac{3V_{0}}{4} and \frac{V_{0}}{4}$ have radius R₁, R₂ ,R₃  and R₄  respectively. Then   R₄

Answer:

Explanation:

Potential at the surface of the charged sphere

$V_{0}=\frac{KQ}{R}$  

 $V=\frac{KQ}{r}$ , $r\geq R$

  $= \frac{KQ}{2R^{3}}(3R^{2}-r^{2});r\leq R$

   $V_{centre}=V_{c}=\frac{KQ}{2R^{3}}\times3R^{2}$

    $\frac{3KQ}{2R}=\frac{3V_{0}}{2}$

   $R_{1}=0$

 As potential  decreases for outside points.

Thus, according to the question, we can write

     $V_{R_{2}}=\frac{5V_{0}}{4}=\frac{KQ}{2R^{3}}(3R^{2}-R^{2}_{2})$

     $\frac{5V_{0}}{4}=\frac{V_{0}}{2R^{2}}( 3R^{2}-R_{2}^{2})$

  or      $\frac{5}{2}=3-\left(\frac{R_{2}}{R}\right)^{2}$

          $R_{2}=\frac{R}{\sqrt{2}}$

  Similarly,

      $V_{R_{3}}=\frac{3V_{0}}{4}$

  $\Rightarrow$    $\frac{KQ}{R_{3}}=\frac{3}{4}\times\frac{KQ}{R}$

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

                    $V_{R_{4}}=\frac{KQ}{R_{4}} =\frac{V_{0}}{4}$

$\Rightarrow$     $\frac{KQ}{R_{4}}=\frac{1}{4}\times\frac{KQ}{R}$

  OR             R₄   =4R