Solution:

Ans. (i) As the electrostatic field inside a conductor is zero, using Gauss’s law, charge on the inner surface of the shell $=-Q$

Charge on the outer surface of the shell $=+\mathrm{Q} \quad 1 / 2$

(ii) To show using Gauss’s law

Expression for electric field for radius, $r=\frac{a}{2}$ :

$$ E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{a^{2}} $$

Expression for electric field for radius, $r=2 b$ :

$$ \begin{equation*} E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{b^{2}} \tag{1} \end{equation*} $$

[CBSE Marking Scheme, 2017]

Detailed Answer :

(i) Charge placed at the centre of a shell is $+Q$. Hence, a charge of magnitude $-Q$ will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is $-Q$.

A charge of $+Q$ is induced on the outer surface of the shell, so a charge of magnitude $Q$ is placed on the outer surface of the shell.

(ii) The electric flux is electric field times the area of spherical surface

$$ \phi=E A=4 E \pi r^{2}=\frac{Q}{\varepsilon_{0}} $$

So electric field outside the sphere $(r>R)$ is seen to be identical to that of a point charge $Q$ at center of sphere.

$$ E=\frac{Q}{4 \pi \varepsilon_{0} r^{2}} $$

Magnitude of electric field vector at a distance

$$ \begin{aligned} r & =\frac{a}{2} \ E & =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{a^{2}} \end{aligned} $$

Magnitude of electric field vector at a distance

$$ \begin{aligned} & r=2 b \ & E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{4 b^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{b^{2}} \mathbf{1} \end{aligned} $$