Q86 / 90 · JEE Main 2022 (27 Jul Shift 2) · Physics

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 J. (Where ϵ0 is the permittivity of free space)

Correct answer: 240

Solution

The figure below shows the parallel plate capacitor with dielectric slab placed in between the plates.

Figure

Now, the effective capacitance of the combination is 

Ceff=ε07×4410+5ε01×4410×10-2

=70ε0+50ε0×10-2=120ε0×10-2

⇒Ceff =1.2ε0

Electrostatics energy of this system is =12CeffV2

=121.2ε02020=240ε0

Hence, the correct answer is 240.

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