Question

A non conducting sphere of radius ${}^{\prime }{a}^{\prime }$ has a type charge ${}^{\prime }+q^{\prime }$ uniformly distributed throughout in volume. A hollow spherical conductor having inner and outer, radii ${}^{\prime }{b}^{\prime }$ and ${}^{\prime }{c}^{\prime }$ and net charge ${}^{\prime }-q^{\prime }$ 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 $=\frac{1}{4\pi {\epsilon }_0}\frac{qr}{a^3}$ for $r<a$.
(ii) The electric field at distance $r$ from the center of the sphere for $a<r<b=0$
(iii) The electric field at distance $r$ from the center of the sphere for $b<r<c<=0$
(iv) The charge on the inner surface of the hollow sphere $=-q$
(v) The charge on the outer surface of the hollow sphere $=+q$.

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

Answer 1

The correct option is B (i),(iii) and (iv)

Solution

Let $q^1$ is the inner surface charge.

then, $q_{outer}=-q-q^1$ such that

$q_{inner}+q_{outer}=-q$

Electric field inside

the conductor should

be zero ( r > b and r < e)

so in the arbitrary

Gaussian surface

Electric flux = $\frac{q_{inside}}{{\epsilon }_0}=0$

$\Rightarrow q_{inside}(b<r<c)=q^1+q=0$

$q^1=-q$

So, $q_{inner}=-q$ and $q_{outer}=0$

(i) Electric field for r < a ia,

$E.4\pi r^2=\frac{q_{inside}}{{\epsilon }_0}$

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 = 0

(iv) Electric field for c < r

$E=\frac{k_{qouter}}{r^2}=k\frac{(0{)}^2}{r^2}=0$

Option - B is correct.