Question
The region between two concentric spheres of radii $$'a'$$ and $$'b',$$ respectively (see figure), have volume charge density $$r = \frac{A}{r},$$ where $$A$$ is a constant and $$r$$ is the distance from the centre. At the centre of the spheres is a point charge $$Q.$$ The value of $$A$$ such that the electric field in the region between the spheres will be constant, is :
The region between two concentric spheres of radii $$'a'$$ and $$'b',$$ respectively (see figure), have volume charge density $$r = \frac{A}{r},$$ where $$A$$ is a constant and $$r$$ is the distance from the centre. At the centre of the spheres is a point charge $$Q.$$ The value of $$A$$ such that the electric field in the region between the spheres will be constant, is :
A.
$$\frac{{2Q}}{{p\left( {{a^2} - {b^2}} \right)}}$$
B.
$$\frac{{2Q}}{{p{a^2}}}$$
C.
$$\frac{Q}{{2p{a^2}}}$$
D.
$$\frac{Q}{{2p\left( {{b^2} - {a^2}} \right)}}$$
Answer :
$$\frac{Q}{{2p{a^2}}}$$
Solution :
Applying Gauss's law
$$\eqalign{ & \oint_S {\overrightarrow E .\overrightarrow {ds} = \frac{Q}{{{ \in _0}}}} \cr & \therefore E \times 4\pi {r^2} = \frac{{Q + 4\pi a{r^2} - 4\pi A{a^2}}}{{{ \in _0}}} \cr & \rho = \frac{{dr}}{{dv}} \cr & Q = {\rho ^4}\pi {r^2} \cr & Q = \int\limits_a {\frac{A}{r}4\pi {r^2}dr = 4\pi A\left[ {{r^2} - {a^2}} \right]} \cr & E = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{Q - 4\pi A{a^2}}}{{{r_0}}} + 4\pi A} \right] \cr} $$
For $$E$$ to be independent of $$'r'$$
$$\eqalign{ & Q - 2\pi A{a^2} = 0 \cr & \therefore A = \frac{Q}{{2\pi {a^2}}} \cr} $$
Applying Gauss's law
$$\eqalign{ & \oint_S {\overrightarrow E .\overrightarrow {ds} = \frac{Q}{{{ \in _0}}}} \cr & \therefore E \times 4\pi {r^2} = \frac{{Q + 4\pi a{r^2} - 4\pi A{a^2}}}{{{ \in _0}}} \cr & \rho = \frac{{dr}}{{dv}} \cr & Q = {\rho ^4}\pi {r^2} \cr & Q = \int\limits_a {\frac{A}{r}4\pi {r^2}dr = 4\pi A\left[ {{r^2} - {a^2}} \right]} \cr & E = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{Q - 4\pi A{a^2}}}{{{r_0}}} + 4\pi A} \right] \cr} $$
For $$E$$ to be independent of $$'r'$$
$$\eqalign{ & Q - 2\pi A{a^2} = 0 \cr & \therefore A = \frac{Q}{{2\pi {a^2}}} \cr} $$
