Question

A solid sphere of radius a has volume charge density p given by $p=p_1,0\le r\le \frac{a}{2}$ $=p_2,\frac{a}{2}\le r\le a$ and electric potential at center is twice as electric potential at surface:

• A. $p_1=2p_2$
• B. $p_1=-p_2$
• C. $p_1=5p_2$
• D. $p_1=-3p_2$

Answer 1

The correct option is C $p_1=5p_2$

$0\le r\le \frac{a}{2}:\rho ={\rho }_1$

$\frac{a}{2}\le r\le a:\rho ={\rho }_2$

The system is equivalent to two solid spheres (i) radius a, density ${\rho }_2$ (ii) radius $\frac{a}{2}$, density $({\rho }_1-{\rho }_2)$.

$\therefore$ potential at $O:V_o=\frac{1}{4\pi {\epsilon }_o}.\frac{3}{2}.\frac{4\pi a^3}{3}\frac{{\rho }_2}{a}+\frac{1}{4\pi {\epsilon }_o}.\frac{3}{2}.\frac{4\pi }{3}\frac{a^3}{8}\times ({\rho }_1-{\rho }_2)\times \frac{2}{a}$

$=\frac{a^2{\rho }_2}{2{\epsilon }_o}+\frac{a^2}{8{\epsilon }_o}({\rho }_1-{\rho }_2)=\frac{a^2}{8{\epsilon }_o}({\rho }_1+3{\rho }_2)$

Potential on the surface at P :

$V_P=\frac{1}{4\pi {\epsilon }_o}.\frac{4\pi a^3}{3}\frac{{\rho }_2}{a}+\frac{1}{4\pi {\epsilon }_o}.\frac{4\pi }{3}\frac{a^3}{8}\frac{({\rho }_1-{\rho }_2)}{a}$

$=\frac{a^2{\rho }_2}{3{\epsilon }_o}+\frac{a^2({\rho }_1-{\rho }_2)}{24{\epsilon }_o}=\frac{a^2}{24{\epsilon }_o}(8{\rho }_2+{\rho }_1-{\rho }_2)$

$\frac{a^2}{24{\epsilon }_o}({\rho }_1+7{\rho }_2)$

$V_o=2V_p$, then, 2$({\rho }_1+7{\rho }_2)=(3{\rho }_1+9{\rho }_2)$

i.e., $2{\rho }_1+14{\rho }_2=3{\rho }_1+9{\rho }_2$

${\rho }^{}=5{\rho }_2$