Question

A parallel plate capacitor of capacitance $90pf$ is connected to a battery of emf $20v$. if a dielectric material of dielectric constant $K$ equal $5/3$ is inserted between the plates the magnitude of the induced charge will be:

• A. $2.4nC$
• B. $0.9nC$
• C. $1.2nC$
• D. $0.3nC$

Answer 1

The correct option is C

$1.2nC$

Step 1: Given Data

Capacitance, $C_0=90pf$

Voltage,$V=20v$

Dielectric constant, $K=5/3$

Step 2: Find the magnitude of the induced charge

Battery remains connected as dielectric is introduced. So $E,V$ unchanged.

Induced charge is

$$\begin{gathered} q=q{q}_0 \\ q=C_{0}V(K-1) \\ q=90x{10}^{-12}x20\times \left(\right(5/3)-1) \\ q=1.2nC \end{gathered}$$

Hence, option C is correct.