Question

Potential difference between centre and the surface of sphere of R and having uniform volume charge density. p with it will be :

• A. $\frac{p{R}^2}{6{ϵ}_0}$
• B. $\frac{p{R}^2}{4{ϵ}_0}$
• C. 0
• D. $\frac{p{R}^2}{2{ϵ}_0}$

Answer 1

The correct option is A $\frac{p{R}^2}{6{ϵ}_0}$

Potential at center, $V_c=\frac{KQ}{2R^2}(3R^2-r^2)$

$\Rightarrow (r-c)V_c=\frac{3KQ}{2R}$

Potential at surface , $V_5\frac{KQ}{R}$

$\Rightarrow △V=\frac{KQ}{2R}=\frac{K}{2R}\times p\times \frac{4}{3}\pi R^3$

$\Rightarrow △V=\frac{p{R}^2}{6{\epsilon }_0}$

Hence, the answer is $△V=\frac{p{R}^2}{6{\epsilon }_0}.$