Question

A parallel plate capacitor is made of two plates of length $I$, width $w$ and separated by distance $d$. A dielectric slab (dielectric constant K) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F=-\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$ then the force on the dielectric when it is near the edge is :

• A. $\frac{Q^{2}d}{2\omega l^2{\in }_0}K$
• B. $\frac{Q^{2}w}{2d{l}^2{\in }_0}\left(K-1\right)$
• C. $\frac{Q^{2}d}{2w{l}^2{\in }_0}\left(K-1\right)$
• D. $\frac{Q^{2}w}{2d{l}^2{\in }_0}K$

Answer 1

Answer: (A)

$C=C_1+C_2$
$=\frac{K\left(x\omega \right){\epsilon }_0}{d}+\frac{\left(1-x\right)\omega {\epsilon }_0}{d}$
$C=\frac{\omega {\epsilon }_0}{d}\left[kx+\left(\ell -x\right)\right]$
$C=\frac{\omega {\epsilon }_0}{d}\left[\ell +\left(k-1\right)x\right]$
$U=\frac{1}{2}\frac{Q^2}{C}=\frac{d.Q^2}{2\omega {\epsilon }_0\left[\ell +\left(k-1\right)x\right]}$
$\frac{du}{dx}=\frac{d.Q^2\left(K-1\right)}{2\omega {\epsilon }_0{\left[\ell +\left(K-1\right)x\right]}^2}$
$F=-\frac{du}{dx}=\frac{d.Q^2\left(K-1\right)}{2\omega {\ell }^2{\epsilon }_0}$ at $\left(x=0\right)$