SATHEE: Electrostatics 5 Question 19

Electrostatics 5 Question 19

21. A parallel plate capacitor having plates of area $S$ and plate separation $d$ , has capacitance $C_{1}$ in air. When two dielectrics of different relative permittivities

$(ε_{1} = 2$ and $ε_{2} = 4)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C_{2}$ . The ratio $\frac{C_{2}}{C_{1}}$ is

(a) $\frac{6}{5}$

(b) $\frac{5}{3}$

(d) $\frac{7}{3}$

(2015 Adv.)

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Solution:

$C_{1} = \frac{ε_{0} s}{d}, C = \frac{2 ε_{0} \frac{s}{2}}{\frac{d}{2}} = \frac{2 ε_{0} s}{d} \Rightarrow C^{'} = \frac{4 ε_{0} \frac{s}{2}}{\frac{d}{2}} = \frac{4 ε_{0} s}{d}$

$\begin{aligned} and C^{''} = \frac{2 ε_{0} \frac{s}{2}}{d} = \frac{ε_{0} s}{d} \\ C_{2} = \frac{C C^{'}}{C + C^{'}} + C^{''} = \frac{4}{3} \frac{ε_{0} s}{d} + \frac{ε_{0} s}{d} = \frac{7}{3} \frac{ε_{0} s}{d} \frac{C_{2}}{C_{1}} = \frac{7}{3} \end{aligned}$

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Electrostatics 5 Question 19

21. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When two dielectrics of different relative permittivities

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21. A parallel plate capacitor having plates of area $S$ and plate separation $d$ , has capacitance $C_{1}$ in air. When two dielectrics of different relative permittivities