A parallel plate capacitor without a dielectric inserted is connected to a batte

Question

A parallel plate capacitor without a dielectric inserted is connected to a battery. The energy density is 14 J/m3 initially. While still connected to the battery, the plates are move closer by a factor of 2.6. Then, the wires are removed from the plates and the distance between the plates is increased by factor of 2.9. After this a dielectric with a dielectric constant of 1.8 is inserted between the sheets. What is the energy density in this final configuration in J/m^3?

(please only answer if you can get the correct answer)

CORRECT ANSWER: 170.35 J/m^3

Explanation / Answer

capacitor is connected with battery so voltage drop will not change.

energy density = U / Volume = (C V^2 / 2 )/ (A d)

and C = e0 er A / d

so energy density = e0 er V^2 / (2 d^2)

d' = d/2.6

so new energy density = (2.6^2) (14) = 94.64 j/m^3

As wire are removed so charge on plates will remain same.

energy density = (Q^2 / 2 C ) / A d

= Q^2 / 2 er e0 A^2  

independent of d hence no change.

after that dielectric will increase the energy density

so energy density will increase by factor of 1.8.

New energy density = 94.64 x 1.8

= 170.35 J / m^3

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