Question

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).

Answer 1

The correct answer is 240
$d_1 = 4×10^{-3}$
$A_1 = 8×4×10^{-4}m^2$
V = 20V
$d_2 = 4×10^{-3},$
$A_2 = 4×1×10^{-4}m^2$
$C_{eq} =\frac{(A_1+5A_2-A_2)_{ε_0}}{d} = \frac{3(16)×10^{-4}}{4×10^{-3}}ε_0$
$ε = \frac{1}{2} C_{eq}V^2$
$= \frac{3}{2}(\frac{4}{10})(400)ε_0 = 240ε_0$
Therefore , the electrostatic energy of this system will be 240 ε0 J