Q 12-02-201JEE MainJEE Main 2018 (8 Apr)Medium
A parallel plate capacitor of capacitance $90$ pF is connected to a battery of emf $20$ V. If a dielectric material of dielectric constant $K = \dfrac53$ is inserted between the plates, the magnitude of the induced charge will be:
Answer: (B) $1.2$ nC
With the battery connected, the free charge becomes
$$Q = KCV = \frac53\times90\times10^{-12}\times20 = 3\times10^{-9}\ \text{C}$$
The induced (bound) charge on the dielectric is
$$Q_i = Q\left(1 - \frac1K\right) = 3\times\frac25 = 1.2\ \text{nC}$$
Solution by Sreeraj P, M.Sc Physics