Electrostatics 5 Question 22
24. A parallel plate capacitor is made of two circular plates separated by a distance of $5 mm$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} V / m$, the charge density of the positive plate will be close to
(a) $6 \times 10^{-7} C / m^{2}$
(b) $3 \times 10^{-7} C / m^{2}$
(c) $3 \times 10^{4} C / m^{2}$
(d) $6 \times 10^{4} C / m^{2}$
(2014 Main)
Show Answer
Solution:
- When free space between parallel plates of capacitor,
$$ E=\frac{\sigma}{\varepsilon _0} $$
When dielectric is introduced between parallel plates of capacitor, $E^{\prime}=\frac{\sigma}{K \varepsilon _0}$
Electric field inside dielectric, $\frac{\sigma}{K \varepsilon _0}=3 \times 10^{4}$
where, $K$ =dielectric constant of medium $=2.2$
$\varepsilon _0=$ permitivity of free space $=8.85 \times 10^{-12}$
$$ \begin{aligned} \Rightarrow \quad \sigma & =2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^{4} \\ & =6.6 \times 8.85 \times 10^{-8}=5.841 \times 10^{-7} \\ & =6 \times 10^{-7} C / m^{2} \end{aligned} $$
