A parallel plate capacitor of plate area A and plate separation d is charged to potential…

Physics · JEE Advanced · NTA ExamsElectrostatics

A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in question, in the process of inserting the slab, then
  1. \(\mathrm{Q}=\frac{\varepsilon_{0} \mathrm{AV}}{\mathrm{~d}}\)
  2. \(\mathrm{Q}=\frac{\varepsilon_{0} \mathrm{KAV}}{\mathrm{~d}}\)
  3. \(\mathrm{E}=\frac{\mathrm{V}}{\mathrm{Kd}}\)
  4. \(\mathrm{W}=\frac{\varepsilon_{0} \mathrm{AV}^{2}}{2 \mathrm{~d}}\left[1-\frac{1}{\mathrm{~K}}\right]\)

Answer

(A) Q = _ 0 AV ~d, (C) E = V Kd, (D) W = _ 0 AV ^ 2 2 ~d [1- 1 ~K ]

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Electrostatics questions · Browse all practice questions