Question

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

• A. 4
• B. 4.0
• C. 4.00

Answer 1

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

Before introducing dielectric slab into the capacitor, electrostatic energy stored in the capacitor is $\frac{1}{2}C{V}^2=\frac{1}{2}\left(200\right)(200{)}^2\times {10}^{-6}=4\text{J}$

As Dielectric slab is introduced in the capacitor, new capacitance is $C^{\prime }=2C=400\mu \text{F}$ and as battery remains connected potential difference across capacitor remains to be $200\text{V}$

Final electrostatic energy is $\frac{1}{2}{C}^{\prime }{V}^2=\frac{1}{2}\left(400\right)(200{)}^2\times {10}^{-6}=8\text{J}$

Change in electrostatic energy is $8-4=4\text{J}$

Ans: $4,4.0,4.00$