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# Parallel plate capacitor

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A parallel-plate capacitor is connected to a battery that maintains potential difference V between the plates. If the plates of the capacitor are pulled farther apart, do the following quantities increase, decrease, or remain the same?

(a) Electric field between the plates;
(b) charge on the plates;
(c) capacitance;
(d) energy stored in the capacitor.

https://brainmass.com/physics/electric-magnetic-fields/parallel-plate-capacitor-23617

## SOLUTION This solution is FREE courtesy of BrainMass!

In all these cases the battery is maintaining a constant voltage V. Increasing d decreases the capacitance as C = epsilon o A/d and epsilon o and A remain the same while d increases.

(a) Electric field between the plates decreases

Since E= V/d and V remains constant while d is increased E will decrease.

(b) charge on the plates Q decreases

Since the capacitance of a parallel-plate capacitor is given by the equation C = epsilon o A/d, increasing the separation d decreases the capacitance. With a smaller value of C and a constant value for V, the charge Q = CV will decrease.

(c) capacitance decreases

Since the capacitance of a parallel-plate capacitor is given by the equation C = epsilon o A/d, increasing the separation d decreases the capacitance.

(d) energy stored in the capacitor decreases

Energy = Â½ Q V. Since V remains the same and Q decreases energy decreases.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!