Question

A parallel plate capacitor is made from two small conducting plates which are separated by a distance d = 6 mm, and for simplicity the space between the plates is void of matter (a vacuum). After allowing the plates to charge, the upper plate acquires an excess of positive charges, while the bottom plate acquires a balanced number of negative charges. The battery is disconnected from the capacitor, and a digital voltmeter is used to measure the electric potential difference between the two plates: ΔV = +1.5 volts. This sets up for a uniform electric field the two plates. Sketch this scenario, and determine the magnitude of the electric field.

Answer 1

The magnitude of the electric field between the plates of the parallel plate capacitor is 0.25 V/mm.

Step-by-step explanation:

A parallel plate capacitor is made from two small conducting plates that are separated by a distance of 6 mm. The upper plate acquires an excess of positive charges, while the bottom plate acquires a balanced number of negative charges. When the battery is disconnected and a digital voltmeter is used to measure the potential difference between the plates, it reads +1.5 volts. In this scenario, the electric field between the plates is uniform.

The magnitude of the electric field can be determined by dividing the potential difference ΔV by the distance between the plates d. In this case, the magnitude of the electric field is ΔV/d = 1.5 V / 6 mm = 0.25 V/mm.