A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A…

A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A dielectric slab (dielectric constant $\mathrm{K}$ ) 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. $\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$
  2. $\frac{\mathrm{Q}^2 \mathrm{~W}}{2 \mathrm{dl}^2 \varepsilon_0}(\mathrm{~K}-1)$
  3. $\frac{\mathrm{Q}^2 \mathrm{~d}}{2 \mathrm{wl}^2 \varepsilon_{\mathrm{o}}}(\mathrm{K}-1)$
  4. $\frac{\mathrm{Q}^2 \mathrm{w}}{2 \mathrm{dl}^2 \varepsilon_{\mathrm{o}}} \mathrm{K}$

Solution

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Asked in: JEE Main 2014 (11 Apr Online)

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