Question

Define intensity of electric field (E) and the potential differences (V) between two points. Derive the relation between them.

Answer 1

Hint: We know that the electric field exists if and only if there is an electric potential difference. Thus, the relation between electric field and electric potential can be generally explained as the electric field is the negative area derivative of electric potential.

Complete step by step answer:
The intensity of electric field or simply the electric field intensity at a particular point, is defined as the electric force experienced by a unit positive charge placed at that point. Electric field intensity is a vector quantity. It is denoted by ‘E’. Electric field is radially outward from a positive charge and radially in towards a negative point charge.
The potential difference between two given points is defined as the amount of external work done in moving a unit positive charge from one point to another, slowly. The potential difference between two points is a scalar quantity. It is denoted by ‘V’.
We know, $E = \dfrac{{kQ}}{{{R^2}}}$ where $k = \dfrac{1}{{4\pi {\varepsilon _o}}}$
Also $V = \dfrac{{kQ}}{R}$
From these two equations we can derive the basic relation between E and V, which is given by:
$E = - \dfrac{{dV}}{{dR}}$ .
The negative sign signifies that electric potential decreases in the direction of electric field.

Note:
So here we have defined and explained the terms and given relationship among them. As we see these quantities are related with each other they play a key role in determination of other useful quantities of electric field. We have studied that like charges repel each other and unlike charges attract each other. Some work is done in moving a charge in the area of another charge.