question_answer

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:       [JEE ONLINE 11-04-2014]

A) \[\frac{{{Q}^{2}}d}{2\omega {{1}^{2}}{{\varepsilon }_{o}}}K\]

B) \[\frac{{{Q}^{2}}\omega }{2d{{1}^{2}}{{\varepsilon }_{o}}}\left( K-1 \right)\]

C) \[\frac{{{Q}^{2}}d}{2w{{1}^{2}}{{\varepsilon }_{o}}}\left( K-1 \right)\]

D) \[\frac{{{Q}^{2}}w}{2d{{1}^{2}}{{\varepsilon }_{o}}}K\]