Question

How would you determine the potential difference between two charged parallel plates that are $0.20\;cm$ apart and have an electrical field strength of $6.0\;V/cm$?

Answer 1

Hint: We know that the voltage drop or just voltage or the potential difference across any two given points in space, under an electrical field is called the voltage or the electrical potential difference. This can be measured using a voltmeter or a multimeter, with SI units voltage.
Formula used:
$V=Ed$

Complete step by step answer:
The electrical potential difference is the amount of work done to move a unit positive charge form from one point in space to infinity, against the electrical forces
Then, it can be expressed mathematically as
$V=Ed$
Here, given that the strength of the electrical field is $6.0\;V/cm$ and the distance between the charged plates, thus the charges is $0.20\;cm$
Then, substituting the values in the formula, we have
$V=6\times 0.2=1.2V$
Thus the amount of work done is moving the charges between the potential is $1.2\;V$

Additional Information:
Since the electric field follows inverse square law, it decreases with increase in the distance between the points of interest. However, we can also form a constant uniform electric field by using two parallel conducting plates. Further, the intensity of the electric field can be increased by adding suitable high dielectric material between them. This combination of parallel plates and dielectric is called a capacitor. A capacitor has the ability to store and discharge charges. This property is also called the capacitance of the capacitor.

Note: The electrical forces acting on the unit positive charge can either be attractive or repulsive depending on the nature of the surrounding charges. However, we must try to break this electrical force, for which work must be done by the unit positive charge.