A parallel plate capacitor of capacitance 200 ,μ {F} is connected to a battery of 200 , {V} . A die

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2. Electric Potential and Capacitance

hard

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.$

A

$400$

B

$0.4$

C

$40$

D

$4$

(JEE MAIN-2021)

Solution

$\Delta {U}=\frac{1}{2}(\Delta {C}) {V}^{2}$

$\Delta {U}=\frac{1}{2}({KC}-{C}) {V}^{2}$

$\Delta {U}=\frac{1}{2}(2-1) {CV}^{2}$

$\Delta {U}=\frac{1}{2} \times 200 \times 10^{-6} \times 200 \times 200$

$\Delta {U}=4 {J}$

Standard 12

Physics

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