# Work done by battery and potential energy of a capacitor

I have a doubt about the work done by a battery and the potential energy of a capacitor?

1- Consider a circuit where the capacitors are connected to the terminals of a battery. Through calculations we get that change in potential energy of a battery is half of the work done by a battery. Since they are not equal, does this imply that work done by a battery is done by a non conservative force? Where is the energy lost to?

2- Now consider a dielectric slab inserted partially into the capacitor that is connected to the terminals of a battery. We want to find the force that keeps the dielectric in equilibrium when it is partially inserted into the parallel plate capacitor. To do this we use the principal of virtual work. We equate work done by the external force that keeps the dielectric in equilibrium in moving the dielectric by a small imaginary distance dR. This must be negative of work done by the battery as net work done is zero, there is no change in kinetic energy. (gravity is neglected for a small distance dR). Here however we take work done by the battery as change in potential energy. (In Solving we equate change in potential energy to work done by external force) Why? Isn't that a contradiction to the previous statement or is the previous statement wrong? Is there an alternate way to solve the given problem?

• First is physics.stackexchange.com/q/187774. The circuit will "ring" until resistance in the wires dissipates the excess energy. – BowlOfRed Jul 11 '15 at 6:55
• In an ideal cct wit no R current is infinite to charge cap. Add any R and work it through and answer will always be right. Low R = high I, higher R = lower I. Capacitor voltage will change exponentially and you'll find energy is conserved - with balance becoming heat in r's. – Russell McMahon Jul 11 '15 at 14:51

## 2 Answers

The battery has to do some work in order to create a static electric field inside a capacitor.The current from the battery has to overtake the resistance in order to flow and to reach the charges into the capacitor.So some heat is generated which causes waste to energy.Here the work is done by the battery to charge the capacitor.But the force of the static electric field inside the capacitor(Coulomb force)is surely conservative.But the force for to do the work-done by the battery is not conservative . The second part of the question is not clear...pls modify it .

As an answer to your second question.

If you released the dielectric it would reach the equilibrium position and then overshoot as it would have kinetic energy eventually stopping on the other side and then returning.
Compare this with the oscillation of a spring-mass system.

So the dielectric would undergo oscillatory motion and if damped it would eventually stop at the equilibrium position with some of the energy supplied by the battery dissipated as heat and the rest stored in the capacitor.