Electrostatics Ques 109
- A parallel plate capacitor is made of two circular plates separated by a distance of $5 \mathrm{~mm}$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} \mathrm{~V} / \mathrm{m}$, the charge density of the positive plate will be close to
(a) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$
(b) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$
(c) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$
(d) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$
(2014 Main)
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Answer:
Correct Answer: 109.(a)
Solution:
Formula:
- 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} \mathrm{C} / \mathrm{m}^{2} \end{aligned} $
BETA
