A parallel plate capacitor of capacitance 200 µF is connected to a battery of 200 V.

Question:

A parallel plate capacitor of capacitance 200 µF 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. 

Solution:

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

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

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

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

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

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