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. $\frac{Q^{2}w}{2dl^{2} \in_{0}} \left(K-1\right)$
• C. $\frac{Q^{2}d}{2wl^{2} \in_{0}} \left(K-1\right)$
• D. $\frac{Q^{2}w}{2dl^{2} \in_{0}} K$

Answer 1

Answer: (A)

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