Question

A non conducting sphere of radius a has a net charge $+q$ uniformly distributed throughout its volume. A spherical conducting shell having inner and outer radii $b$ and $c$ and net charge $-q$ is concentric with the sphere (see the figure).

Read the following statements
(i) The electric field at a distance $r$ from the center of the sphere for $r<a$ is $\frac{1}{4\pi {\epsilon }_0}\frac{qr}{a^3}$
(ii) The electric field at distance $r$ for $a<r<b$ is $0$
(iii) The electric field at distance $r$ for $b<r<c$ is $0$
(iv) The charge on the inner surface of the spherical shell is $-q$
(v) The charge on the outer surface of the spherical shell is $+q$
Which of the above statements are true?

• A. $(i)$, $(ii)$ and $(v)$
• B. $(i)$, $(iii)$ and $(iv)$
• C. $(ii)$, $(iii)$ and $(iv)$
• D. $(ii)$, $(iii)$ and $(v)$

Answer 1

Answer: (B)

Let $q^1$ is the inner surface charge.
then, $q_{\text{outer }}=-q-q^1$ such that
$q_{\text{inner }}+q_{\text{outer }}=-q$
Electric field inside
the conductor should
be zero $(r>b$ and $r<e)$
so in the arbitrary
Gaussian surface
Electric flux $=\frac{\text{ qinside }}{{\epsilon }_0}=0$
$\Rightarrow q_{\text{inside }}(b<r<c)=q^1+q=0$
$q^1=-q$
So, $q_{\text{inner }}=-q$ and $q_{\text{outer }}=0$
(i) Electric field for $r<$ a ia,
$E.4\pi r^2=\frac{q_{\text{inside }}}{{\epsilon }_0}$
As a charge is uniformly distributed
$q$ inside $=\left(\frac{r^3}{a^3}\right)q$
So, $E.4\pi r^2=\frac{r^{3}q}{a^3{\epsilon }_0}$
$\Rightarrow E=\frac{1}{4\pi }\times \frac{q}{a^3{\epsilon }_0}$
(ii) Electric field for $a<r<b$,
$E=\frac{kq}{r^2}$
(iii) Electric field for $b<r<c$
As told earlier, $E=O$
(iv) Electric field for $c<r$
$E=\frac{k_{\text{qouter }}}{r^2}=k\frac{(0{)}^2}{r^2}=0$