A parallel plate capacitor of capacitance 200 | vedclass

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(D) Initial capacitance $C_i = 200 \,\mu F = 200 	imes 10^{-6} \,F$.Initial voltage $V = 200 \,V$.Initial electrostatic energy $U_i = \frac{1}{2} C_i

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$4$

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(D) Initial capacitance $C_i = 200 \,\mu F = 200 	imes 10^{-6} \,F$.Initial voltage $V = 200 \,V$.Initial electrostatic energy $U_i = \frac{1}{2} C_i V^2$.When a dielectric slab of constant $K = 2$ is inserted while the battery remains connected,the new capacitance becomes $C_f = K C_i = 2 	imes 200 \,\mu F = 400 \,\mu F$.The voltage $V$ remains constant at $200 \,V$.Final electrostatic energy $U_f = \frac{1}{2} C_f V^2$.The change in electrostatic energy is $\Delta U = U_f - U_i = \frac{1}{2} (C_f - C_i) V^2$.$\Delta U = \frac{1}{2} (K C_i - C_i) V^2 = \frac{1}{2} (K - 1) C_i V^2$.Substituting the values: $\Delta U = \frac{1}{2} (2 - 1) 	imes (200 	imes 10^{-6}) 	imes (200)^2$.$\Delta U = \frac{1}{2} 	imes 1 	imes 200 	imes 10^{-6} 	imes 40000$.$\Delta U = 100 	imes 10^{-6} 	imes 40000 = 4 \,J$.

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