Question

A parallel- plate capacitor is placed in such a way that its plates are horizontal and lower plate is dipped into a liquid of dielectric constant $k$ and density $\rho$. Charge on plates of capacitor is $Q$ and $-Q$. [$+Q$ is charge on upper plate]

• A. Level of liquid in the space between plates rises.
• B. Level of liquid in the space between plates falls.
• C. Net charge on the lower plate is $\frac{-Q}{k}$.
• D. Charge induced on the upper surface of liquid is $-Q(1-k)$

Answer 1

Answer: (A), (C)

The correct options are
A Level of liquid in the space between plates rises.
C Net charge on the lower plate is $\frac{-Q}{k}$.

Due to negative charge induced on upper surface of liquid level of fluid rise.
Induced charge is $-q(1-\frac{1}{k})$