Question

A parallel plate capacitor is made of two plates of length \(l\), 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. \(\frac{Q^2d}{2wl^2\varepsilon_0}K\)
• B. \(\frac{Q^2W}{2dl^2\varepsilon_0}(K-1)\)
• C. \(\frac{Q^2d}{2wl^2\varepsilon_0}(K-1)\)
• D. \(\frac{Q^2w}{2dl^2\varepsilon_0}K\)

Hint:

Treat the partly filled capacitor as two parallel capacitors, find \(U=Q^2/2C\) as a function of x and take \(-dU/dx\) at \(x=0\).

Answer 1

The Correct Option is C

With the slab inside up to distance \(x\), the capacitor is two capacitors in parallel:
\(C=\frac{\varepsilon_0w(l-x)}{d}+\frac{K\varepsilon_0wx}{d}=\frac{\varepsilon_0w}{d}\left[l+(K-1)x\right]\).
At constant charge, \(U=\frac{Q^2}{2C}\), so \(F=-\frac{\partial U}{\partial x}=\frac{Q^2}{2C^2}\frac{dC}{dx}\).
\(\frac{dC}{dx}=\frac{\varepsilon_0w(K-1)}{d}\). Near the edge \(x\to0\): \(C=\frac{\varepsilon_0wl}{d}\).
\(F=\frac{Q^2}{2}\cdot\frac{d^2}{\varepsilon_0^2w^2l^2}\cdot\frac{\varepsilon_0w(K-1)}{d}=\frac{Q^2d(K-1)}{2\varepsilon_0wl^2}\).
Answer: (3) \(\frac{Q^2d}{2wl^2\varepsilon_0}(K-1)\)