A parallel plate capacitor with air between the plates has a capacitance of 9 pF . The separation between the plates is d . The space between the plates is now filled with two dielectrics constant K 1

Question

A parallel plate capacitor with air between the plates has a capacitance of 9 pF . The separation between the plates is d . The space between the plates is now filled with two dielectrics constant K 1 = 3 and thickness d 3 while the other one has dielectric constant K 2 = 6 and thickness 2 d 3 . Capacitance of the capacitor is now

TestHub · Accepted Answer

The two capacitors formed are in parallel, hence capacitance of the combination

$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$ $. . . (i)$

where, $C_{1} = \frac{K_{1} ε_{0} A}{\frac{d}{3}}$ $(∵ C = \frac{ε_{0} A}{d})$ $. . . (ii)$

$C_{2} = \frac{K_{2} ε_{0} A}{\frac{2 d}{3}}$ $. . . (iii)$

$C_{eq} = \frac{\frac{3 K_{1} ε_{0} A}{d} \times \frac{K_{2} ε_{0} A \cdot 3}{2 d}}{\frac{3 K_{1} ε_{0} A}{d} + \frac{3 K_{2} ε_{0} A}{2 d}}$

$= \frac{9 K_{1} K_{2} ε_{0}^{2} A^{2}}{2 d^{2}} \times \frac{d}{ε_{0} A \times 18}$

It is given

$\frac{ε_{0} A}{d} = 9 pF$

Using given values

$C_{eq} = 40.5 pF$

Physics - Capacitor Question with Solution | TestHub

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