Capacitors

JEE Main level, numerical answer, medium

134 of 180

Question 134 of 180

A parallel plate capacitor of capacitance 200μF200\mu F is connected to a battery of 200V.200V. A dielectric slab of dielectric constant 22 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.J.

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Answer: 4

Initially
C=200μFC=200\mu F
Ei=12CV2=12×(200×10−6)×(200)2{E}_{i}=\dfrac{1}{2}C{V}^{2}=\dfrac{1}{2}\times \left(200\times {10}^{-6}\right)\times {\left(200\right)}^{2}
Finally
C′=KC=400μF{C}^{'}=KC=400\mu F
Ef=12C′V2=12×(400×10−6)×(200)2{E}_{f}=\dfrac{1}{2}{C}^{'}{V}^{2}=\dfrac{1}{2}\times \left(400\times {10}^{-6}\right)\times (200{)}^{2}
ΔE=12×[(400−200)×10−6]×4×104=4J\Delta E=\dfrac{1}{2}\times \left[\left(400-200\right)\times {10}^{-6}\right]\times 4\times {10}^{4}=4J

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