Question
Let $$P\left( r \right) = \frac{Q}{{\pi {R^4}}}r$$ be the charge density distribution for a solid sphere of radius $$R$$ and total charge $$Q.$$ For a point $$'p'$$ inside the sphere at distance $${r_1}$$ from the centre of the sphere, the magnitude of electric field is :
A.
$$\frac{Q}{{4\pi { \in _0}r_1^2}}$$
B.
$$\frac{{Qr_1^2}}{{4\pi { \in _0}{R^4}}}$$
C.
$$\frac{{Qr_1^2}}{{3\pi { \in _0}{R^4}}}$$
D.
0
Answer :
$$\frac{{Qr_1^2}}{{4\pi { \in _0}{R^4}}}$$
Solution :
Let us consider a spherical shell of thickness $$dx$$ and radius $$x.$$ The volume of this spherical shell $$ = 4\pi {r^2}dr.$$
The charge enclosed within shell $$ = \frac{{Qr}}{{\pi {R^4}}}\left[ {4\pi {r^2}dr} \right]$$
The charge enclosed in a sphere of radius $${r_1}$$ is $$ = \frac{{4Q}}{{{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr = \frac{{4Q}}{{{R^4}}}\left[ {\frac{{{r^4}}}{4}} \right]_0^{{r_1}} = \frac{Q}{{{R^4}}}r_1^4$$
$$\therefore $$ The electric field at point $$p$$ inside the sphere at a distance $${r_1}$$ from the centre of the sphere is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\left[ {\frac{Q}{{{R^4}}}r_1^4} \right]}}{{r_1^2}} = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{R^4}}}r_1^2$$
Let us consider a spherical shell of thickness $$dx$$ and radius $$x.$$ The volume of this spherical shell $$ = 4\pi {r^2}dr.$$
The charge enclosed within shell $$ = \frac{{Qr}}{{\pi {R^4}}}\left[ {4\pi {r^2}dr} \right]$$
The charge enclosed in a sphere of radius $${r_1}$$ is $$ = \frac{{4Q}}{{{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr = \frac{{4Q}}{{{R^4}}}\left[ {\frac{{{r^4}}}{4}} \right]_0^{{r_1}} = \frac{Q}{{{R^4}}}r_1^4$$
$$\therefore $$ The electric field at point $$p$$ inside the sphere at a distance $${r_1}$$ from the centre of the sphere is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\left[ {\frac{Q}{{{R^4}}}r_1^4} \right]}}{{r_1^2}} = \frac{1}{{4\pi { \in _0}}}\frac{Q}{{{R^4}}}r_1^2$$
