Question

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density $\rho$ within it will be :

• A. $\frac{\rho R^2}{6{\epsilon }_0}$
• B. $\frac{\rho R^2}{4{\epsilon }_0}$
• C. $\frac{\rho R^2}{3{\epsilon }_0}$
• D. $\frac{\rho R^2}{2{\epsilon }_0}$

Answer 1

Answer: (A)

$\frac{\rho R^2}{6{\epsilon }_0}$

Using Gauss's law the electric field inside the sphere , $E.4\pi r^2=\frac{q}{{ϵ}_0}=\rho \frac{\left(4/3\right)\pi r^3}{{ϵ}_0}$

or $E=\frac{\rho r}{3{ϵ}_0}$

Potential difference between surface and center is $V=-{\int }_R^{0}E.dr=-{\int }_R^0\frac{\rho r}{3{ϵ}_0}dr=\frac{\rho R^2}{6{ϵ}_0}$