Question 149: Electrostatics & Capacitance

Question

A parallel plate capacitor of capacitance 200 $$\mu$$F is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____________J.

Accepted Answer

$$\Delta U = {1 \over 2}(\Delta C){V^2}$$
$$\Delta U = {1 \over 2}(KC - C){V^2}$$
$$\Delta U = {1 \over 2}(2 - 1)C{V^2}$$
$$\Delta U = {1 \over 2} 	imes 200 	imes {10^{ - 6}} 	imes 200 	imes 200$$
$$\Delta U = 4$$ J

Electrostatics & Capacitance

Integer / Numeric Answer Type