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 [2015]

Question

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(4)

As we know, the capacitance of a parallel plate capacitor is $d$

Initially, capacitance, $C_{1} = \frac{ε_{0} S}{d}$

When two dielectrics of permittivities $ε_{1} = 2$ and $ε_{2} = 4$ are introduced between the plates, then

$C_{2} = \frac{C_{A} \times C_{B}}{C_{A} + C_{B}} + C_{C}$ $= \frac{(\frac{2 ε_{0} S}{d}) (\frac{4 ε_{0} S}{d})}{\frac{6 ε_{0} S}{d}} + \frac{ε_{0} S}{d}$

$= \frac{4}{3} \frac{ε_{0} S}{d} + \frac{ε_{0} S}{d}$

$∴$ $C_{2} = \frac{7}{3} \frac{ε_{0} S}{d} = \frac{7}{3} C_{1}$ $= \frac{C_{2}}{C_{1}} = \frac{7}{3}$ $[∵ C_{1} = \frac{ε_{0} S}{d}]$

Solution

1 $\frac{6}{5}$

2 $\frac{5}{3}$

3 $\frac{7}{5}$

4 $\frac{7}{3}$

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Solution

Q. 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 (ε1=2 and ε2=4) are introduced between the two plates as shown in the figure, the capacitance becomes C2. The ratio C2C1 is [2015]

1 65

2 53

3 75

4 73

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1 $\frac{6}{5}$

2 $\frac{5}{3}$

3 $\frac{7}{5}$

4 $\frac{7}{3}$