Capacitors

JEE Main level, single correct, medium

131 of 180

Question 131 of 180

A parallel plate capacitor filled with a medium of dielectric constant 1010, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 1515. Then the energy of capacitor will

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Answer: (A) increase by 5050%

When a dielectric of dielectric constant KK is inserted between the plates of a capacitor C0{C}_{0}, the capacitance becomes KC0K{C}_{0}.
Energy stored in a capacitor is given by, U=12CV2U=\dfrac{1}{2}C{V}^{2}. Now,
Ui=12(K1C0)V2{U}_{i}=\dfrac{1}{2}\left({K}_{1}{C}_{0}\right){V}^{2} and Uf=12(K2C0)V2{U}_{f}=\dfrac{1}{2}\left({K}_{2}{C}_{0}\right){V}^{2}
⇒ΔU=Uf−Ui=12(K2−K1)C0V2\Rightarrow \Delta U={U}_{f}-{U}_{i}=\dfrac{1}{2}\left({K}_{2}-{K}_{1}\right){C}_{0}{V}^{2}
Percentage change in the energy will be,
ΔUUi×100=12×5×C0212×10×C02×100=50\dfrac{\Delta U}{{U}_{i}}\times 100=\dfrac{\dfrac{1}{2}\times 5\times {{C}_{0}}^{2}}{\dfrac{1}{2}\times 10\times {{C}_{0}}^{2}}\times 100=50%

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