Question 140: Electrostats & Capacitance

Question

A parallel-plate capacitor of capacitance $40 \mu \mathrm{~F}$ is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $\mathrm{K}=2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

Accepted Answer

Given, $$K = 2,\,C = 40\mu F,\,V = 100\,V$$
$$\Delta q = (KC)V - CV$$ (As $$q = CV$$)
$$ = (K - 1)CV$$
$$ = (2 - 1) 	imes 40 	imes {10^{ - 6}} 	imes 100 = 4 	imes {10^{ - 3}}C$$
$$\Delta q = 4\,mC$$
$$\Delta u = {1 \over 2}C'{V^2} - {1 \over 2}C{V^2}$$
$$ = {1 \over 2}{V^2}(C' - C) = {1 \over 2}{V^2}(KC - C)$$
$$ = {1 \over 2}{V^2}C(K - 1) = {1 \over 2} 	imes {10^4} 	imes 40 	imes {10^{ - 6}}(2 - 1)$$
$$ = \Delta u = 2 	imes {10^{ - 1}} = 0.2\,J$$
$$ \Rightarrow \Delta u = 0.2\,J$$

Answer

8 mC and 2.0 J

Answer

4 mC and 0.2 J

Answer

2 mC and 0.2 J

Answer

2 mC and 0.4 J

Electrostats & Capacitance

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