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Answer please! A parallel plate capacitor is made of two square plates of side 'a' , separated by a distance d (d>>a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of

A parallel plate capacitor is made of two square plates of side 'a' , separated by a distance d (d>>a). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is :

• Option 1)

$\frac{K\epsilon _{0}a^{2}}{2d(K+1)}\; \;$

• Option 2)

$\; \frac{K\epsilon _{0}a^{2}}{d(K-1)}1n\, K\; \;$

• Option 3)

$\;\frac{K\epsilon _{0}a^{2}}{d}1n\, K\; \;$

• Option 4)

$\;\frac{1}{2}\frac{K\epsilon _{0}a^{2}}{d}\; \; \;$

Answers (1)
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If K filled between the plates -

$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$

- wherein

$\dpi{100} C\propto A$

$\dpi{100} C\propto\frac{1}{d}$

$\frac{y}{x} = \frac{d}{a}$

$dy = \frac{d}{a} dx$

$dC_{1} = \frac{\varepsilon_{o}adx}{d-y} dx$

$dC_{2} = \frac{K\varepsilon_{o}adx}{y}$

$dC_{eq} =\frac{dC_{1}dC_{2}}{dC_{1}+dC_{2}} = \frac{K\varepsilon_{o}adx}{Kd + (1-K)y}$

$C_{eq} = \int_{0}^{a} \frac{K\varepsilon_{o}adx}{Kd + (1-K)y}$

$= \frac{K\varepsilon_{o}a^{2}ln K}{ d(K - 1)}$

Option 1)

$\frac{K\epsilon _{0}a^{2}}{2d(K+1)}\; \;$

Option 2)

$\; \frac{K\epsilon _{0}a^{2}}{d(K-1)}1n\, K\; \;$

Option 3)

$\;\frac{K\epsilon _{0}a^{2}}{d}1n\, K\; \;$

Option 4)

$\;\frac{1}{2}\frac{K\epsilon _{0}a^{2}}{d}\; \; \;$

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