Question

A parallel plate capacitors of capacitance $C_0$ is charged to voltage $V$ and then battery is disconnected. A dielectric covering one third area of each plate is now inserted. Charges on plates get redistributed such that portion covered by dielectric and vaccum shares equal amount of charge, then which of the statements are true?

• A. Dielectric constant is 2.
• B. Charges due to polarization on surface of dielectric is 0.25 $C_{0}V$.
• C. Electric field inside dielectric is same as in vaccum.
• D. Electric field inside dielectric is $\frac{2}{3}$ of that in vaccum.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Dielectric constant is 2.
B Charges due to polarization on surface of dielectric is 0.25 $C_{0}V$.
C Electric field inside dielectric is same as in vaccum.

$Q_0=C_{0}V$

$△V_1=△V_2$
$\frac{Q_0}{2}\frac{d}{{\epsilon }_0\frac{A}{3}k}=\frac{Q_0}{2}\frac{d}{{\epsilon }_0\frac{2A}{3}}\Rightarrow k=2$

$Q_p=Q(1-\frac{1}{k})=\frac{Q}{2}$ where $Q=\frac{Q_0}{2}$
$\therefore Q_P=\frac{Q_0}{4}=0.25C_{0}V$

$△V_1=△V_2\Rightarrow E_{1}d=E_{2}d$