JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • 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 Main Online Paper ( Held On 11 Apirl  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) \[\frac{{{Q}^{2}}w}{2d{{1}^{2}}{{\varepsilon }_{o}}}K\]

    Correct Answer:


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