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A parallel plate capacitor is made of two circular plates separated by a distance of $5mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $ 3 \times 10^4 \;V/m$, the change density of the positive plate will be close to

$\begin{array}{1 1}(A)\;3 \times 10^4 \;C/m^2 \\ (B)\;6 \times 10^4 \;C/m^2 \\(C)\;6 \times 10^{-7} \;C/m^2 \\(D)\;3 \times 10^{-7} \;C/m^2 \end{array} $

1 Answer

The change density of the positive plate will be close to $6 \times 10^{-7} \;C/m^2$
Hence C is the correct answer.
answered Apr 7, 2014 by meena.p
 
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