electric-charges-and-fields Question 49
Question: Q. 8. A point charge $+Q$ is placed at the centre $O$ of an uncharged hollow spherical conductor of inner radius ’ $a$ ’ and outer radius ’ $b$ ‘. Find the following:
(i) The magnitude and sign of the charge induced on the inner and outer surface of the conducting shell.
(ii) The magnitude of electric field vector at a distance (i) $r=\frac{a}{2}$, and (ii) $r=2 b$, from the centre of the shell.
A&E [CBSE SQP 2017-18]
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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
$$ \begin{aligned} & r=2 b \ & E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{4 b^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{b^{2}} \mathbf{1} \end{aligned} $$
