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A parallel-plate capacitor has plate area A. A battery is used to charge the capacitor so that the magnitude of charge on each plate is Q, and then is disconnected. Initially, the capacitor has a plate separation of d. At this separation the capacitor contains energy U. The plates are then moved to a separation of 2d without disturbing the charge. What is the energy of the capacitor at this larger plate separation?

Respuesta :

jorgezarate

Answer:

[tex]U_f=2U_o[/tex]

When we double the separation between the plates of the capacitor the electric Energy is doubled too.

Explanation:

First we analyse the Capacitance.

Initial: [tex]C_o=\epsilon*S/d[/tex]

Final:  [tex]C_f=\epsilon*S/(2d)=C_o/2[/tex]

Energy:

Initial: [tex]U_o=Q^2/(2*C_o)[/tex]

Initial: [tex]U_f=Q^2/(2*C_f)=2*Q^2/(2*C_o)=2U_o[/tex]

When we double the separation between the plates of the capacitor the Energy electric is doubled too.

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