A parallel plate capacitor is made of two plates of length l , width ω and separated by…
Physics · JEE Main · NTA Exams — Electrostatics
A parallel plate capacitor is made of two plates of length l, width ω 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 = –\(\frac{\partial \mathrm{U}}{\partial \mathrm{x}}\)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:
- \(\frac{\mathrm{Q}^{2} \omega}{2 \mathrm{~d} l^{2} \epsilon_{0}}(\mathrm{k}-1)\)
- \(\frac{\mathrm{Q}^{2} \omega}{2 \mathrm{~d} l^{2} \epsilon_{0}} \mathrm{k}\)
- \(\frac{\mathrm{Q}^{2} \mathrm{~d}}{2 \omega l^{2} \epsilon_{0}}(\mathrm{k}-1)\)
- None
Answer
(C) Q ^ 2 ~d 2 l^ 2 _ 0 ( k -1)
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