An air-filled parallel plate capacitor is charged by a battery so that it stores an electrostatic energy U₀ . The battery is then disconnected. A dielectric slab of dielectric constant K = 2 and thickness equal to half the plate separation ( d 2 ) is inserted between the plates. The new electrostatic energy stored in the capacitor will be:

Step-by-step solution

Let the initial capacitance be C₀ = ₀A d . The initial energy stored is U₀ = Q² 2C₀ , where Q is the charge on the plates. When the battery is disconnected, the charge Q remains constant. A dielectric slab of thickness t = d 2 and K = 2 is inserted. The new capacitance C is: C = ₀A d - t + t K = ₀A d - d 2 + d 4 C = ₀A 3d 4 = 4 3 C₀ The new energy stored in the capacitor is: U = Q² 2C = Q² 2 ( 4 3 C₀ ) U = 3 4 ( Q² 2C₀ ) = 3 4 U₀ Answer: 3 4 U₀