# 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 U F 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:

Question:

A parallel plate capacitor is made of two plates of length 1 , 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 $\mathrm{F}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}$ where $\mathrm{U}$ 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:

1. $\frac{\mathrm{Q}^{2} \mathrm{~d}}{2 \omega l^{2} \varepsilon_{\mathrm{o}}} \mathrm{K}$

2. $\frac{\mathrm{Q}^{2} \omega}{2 \mathrm{~d} l^{2} \varepsilon_{0}}(\mathrm{~K}-1)$

3. $\frac{\mathrm{Q}^{2} \mathrm{~d}}{2 \mathrm{w} l^{2} \varepsilon_{0}}(\mathrm{~K}-1)$

4. $\frac{\mathrm{Q}^{2} \mathrm{w}}{2 \mathrm{~d} l^{2} \varepsilon_{0}} \mathrm{~K}$

Correct Option: 3

Solution:

(3)

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