Electrostatics Ques 102
- A parallel plate capacitor of capacitance $90 \mathrm{pF}$ is connected to a battery of emf $20 \mathrm{~V}$. If a dielectric material of dielectric constant $K=5 / 3$ is inserted between the plates, the magnitude of the induced charge will be
(2018 Main)
(a) $0.9 \mu \mathrm{C}$
(b) $1.2 \mu \mathrm{C}$
(c) $0.3 \mu \mathrm{C}$
(d) $2.4 \mu \mathrm{C}$
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Answer:
Correct Answer: 102.(b)
Solution:
Formula:
$ \begin{aligned} Q & =K C_{0} V \\ \left|Q_{\text {in }}\right| & =\left|Q \quad 1-\frac{1}{K}\right| \\ & =\left(90 \times 10^{-12}\right)(20) \frac{5}{3} \quad 1-\frac{3}{5} \mathrm{C} \\ & =1200 \times 10^{-12} \mathrm{C}=1.2 \mu \mathrm{C} \end{aligned} $
BETA
