Two square-shaped metal plates of side 1 ,m, | vedclass

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(C) The capacitance $C$ of the parallel plate capacitor as it is immersed in oil to a depth $x$ is given by the sum of the capacitances of the part in

Vedclass · Accepted Answer

$4.425 	imes 10^{-9} \,A$

Vedclass · Answer

(C) The capacitance $C$ of the parallel plate capacitor as it is immersed in oil to a depth $x$ is given by the sum of the capacitances of the part in air and the part in oil:$C = \frac{\epsilon_0 (A - Ax)}{d} + \frac{K \epsilon_0 Ax}{d} = \frac{\epsilon_0 A}{d} [1 + (K - 1)x/L]$Given side $L = 1 \,m$, so $A = L^2 = 1 \,m^2$. The depth $x$ increases with time $t$ as $x = vt$, where $v = 0.001 \,m/s$.$C(t) = \frac{\epsilon_0}{d} [1 + (K - 1)vt]$Differentiating with respect to time $t$:$\frac{dC}{dt} = \frac{\epsilon_0}{d} (K - 1) v$Substituting the values: $\epsilon_0 = 8.85 	imes 10^{-12} \,F/m$, $d = 0.01 \,m$, $K = 11$, $v = 0.001 \,m/s$:$\frac{dC}{dt} = \frac{8.85 	imes 10^{-12}}{0.01} 	imes (11 - 1) 	imes 0.001 = 8.85 	imes 10^{-12} 	imes 10 	imes 0.1 = 8.85 	imes 10^{-12} \,F/s$The current $I$ drawn from the battery is $I = V \frac{dC}{dt}$:$I = 500 	imes 8.85 	imes 10^{-12} = 4.425 	imes 10^{-9} \,A$.

Similar Questions

$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

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