An air spaced parallel plate capacitor has an initial charge of 0.05 uC after be

Question

An air spaced parallel plate capacitor has an initial charge of 0.05 uC after being connected to a 10 V battery. (A) What is the total energy stored between the plates of the capacitor? (B) If the battery is disconnected and the plate separation is tripled to 0.3 mm, what is the electric field before and after the plate separation change? (C) What is the final voltage across the plates and the final energy stored? (D) Calculate the work done in pulling the plates apart. Does this fully account for the energy change in part (B)?

Explanation / Answer

A) Total energy stored =0.5 CV^2= 2.5 uJ B) Charge will remain same Q=0.05*10 uC Cnew=0.05/3 uF SO voltage = Q/Cnew=.30 v Electric Field= V/d=100*10^3 V/m C) Final energy = Q^2/2C=7.5 uJ D) Work done= 7.5-2.5 uJ=5 uJ

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