Question

The region between two concentric spheres of radii ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ , respectively( see figure), has volume charge density $\rho =\frac{A}{r}$, where $A$ is a constant and $r$ is the distance from the centre. At the center 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. $\frac{Q}{2\pi (b^2-a^2)}$
• B. $\frac{2Q}{\pi (a^2-b^2)}$
• C. $\frac{2Q}{\pi a^2}$
• D. $\frac{Q}{2\pi a^2}$

Answer 1

The correct option is D $\frac{Q}{2\pi a^2}$

Radii $=a$ and $b$

charge density $\rho =\frac{A}{r}$

$A=$ constant

$r=$ distance from the centre at the centre of the spheres is a point

charge $=Q$

By symmetry electric field in the region between two shells is radial. Consider a shell of radius $r$. Flux through this shell is

$\phi =\varphi \overline{E}d\overline{s}$

$\phi =E.4\pi r^2=\frac{q_{inc}}{{ϵ}_0}$

Net charge between $r=a$ to $r=r$ would be

$Q={\int }_a^r\rho \left(4\pi r^2\right)dr$

$={\int }_a^r\frac{c}{r}4\pi r^{2}dr$

$=\frac{4\pi C}{2}{\left[r^2\right]}_a^r$

$Q=2\pi C(r^2-a^2)$

$E.4\pi r^2=\frac{Q+q}{{ϵ}_0}$

$E=\frac{K\left(2\pi C\right)(r^2-a^2)+q}{r^2}$

$E=2\pi KC+\frac{K}{r^2}(-2\pi C{a}^2+q)$

Since $E$ be a constant, it should be independent of $R$

$Q-2\pi C{a}^2=6$

$C=\frac{Q}{2\pi a^2}$