Question 536: Electrostatics & Capacitance

Question

Let $$P\left( r \right) = {Q \over {\pi {R^4}}}r$$ be the change 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 center of the sphere, the magnitude of electric field is :

Accepted Answer

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
$$ = {{{Q_r}} \over {\pi {R^4}}}\left[ {4\pi {r^2}dr} ight]$$
The charge enclosed in a sphere of radius $${r_1}$$ is
$${{4Q} \over {{R^4}}}\int\limits_0^{{r_1}} {{r^3}} dr$$
$$ = {{4Q} \over {{R^4}}}\left[ {{{{r^4}} \over 4}} ight]_0^{{r_1}}$$
$$ = {Q \over {{R^4}}}r_1^4$$
$$	herefore$$ The electric field at point $$p$$ inside the sphere at a distance $${r_1}$$ from the center of the sphere is
$$E = {1 \over {4\pi { \in _0}}}{{\left[ {{Q \over {{R^4}}}r_1^4} ight]} \over {r_1^2}}$$
$$ = {1 \over {4\pi { \in _0}}}{Q \over {{R^4}}}r_1^2$$

Answer

$${Q \over {4\pi \,{ \in _0}\,r_1^2}}$$

Answer

$${{Qr_1^2} \over {4\pi \,{ \in _0}\,{R^4}}}$$

Answer

$${{Qr_1^2} \over {3\pi \,{ \in _0}\,{R^4}}}$$

Answer

$$0$$

Electrostatics & Capacitance

Let $P\left( r \right) = {Q \over {\pi {R^4}}}r be the change density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance {r_1}$ from the center of the sphere, the magnitude of electric field is :

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