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A parallel plate capacitor of area $50\: cm^2$ and plate separation $3\: mm$ is charged initially to $80\: \mu C$. Due to a radioactive source nearby, the medium between the plates gets slightly conducting and the plate loses charge initially at the rate of $1.5 \times 10^{-8}C/s$. What is the magnitude and direction of displacement current?

$\begin {array} {1 1} (a)\;\text{Zero} \\ (b)\;1.5 \times 10^{-10}A\text{ and perpendicular to conduction current} \\ (c)\;1.5 \times 10^{-10} \text{ and same direction as conduction current} \\ (d)\;1.5 \times 10^{-8} \text{ and opposite direction to that of conduction current} \end {array}$


1 Answer

Due to leaking, there is flow of +ve charge from +ve plate to –ve plate.
Thus the conduction current within the plates is from +ve plate to –ve plate. Now, the displacement current is
$I_D = \in_oA \times \large\frac{1}{\in_oA} \times \large\frac{dQ}{dt} = \large\frac{dQ}{dt}$
Since, $\large\frac{dQ}{dt}, \large\frac{dE}{dt} < 0. \: I_D$ is opposite to the direction of electric field.
Thus, $ I_D$ has same magnitude as conduction current but opposite direction.
Ans : (d)
answered Feb 27, 2014 by thanvigandhi_1

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