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← JEE Main 2022 · 27 Jul · Shift 2
PhysicsInteger+4 / -1Question 56 of 90

JEE Main 2022 · 27 Jul · Shift 2, Physics question 56

A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1 cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ______ ε0\varepsilon_0 J.

(Where ε0\varepsilon_0 is the permittivity of free space)

Answer and worked solution

Official answer: 240

Parallel parts. C=ε0(0.04)(5×0.01+0.07)0.004=1.2ε0C=\frac{\varepsilon_0(0.04)(5\times0.01+0.07)}{0.004}=1.2\varepsilon_0.

Energy 12CV2=0.6ε0×400=240ε0\frac12CV^2=0.6\varepsilon_0\times400=240\varepsilon_0 J.

Answer: 240