Question

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

Answer 1

The electric field intensity is the measure of the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge which is placed at that point.
The electric field intensity between two points is the vector sum of all the electric fields acting at that point.
The potential difference between two points tells us the amount of energy acquired by a unit charge when moved from one point to the other.

For eg, if the electric p.d. between two points is 5 Volts then a unit charge that is 1 coulomb of charge will acquire 5 joules of potential energy when moved between those two locations.
Derivation:-
We have one formula of energy from the definition of work done $W=F\times d$ ..(1)
The formula of potential difference from its definition above is $V=\frac{energy}{q}$ ..(2)
From coulomb's law we have the formula of force as: $F=qE$ ..(3)
Substituting the value of energy from equation 1 into equation 3 we get, $V=\frac{F\times d}{q}$ ..(4)
Substituting the value of $F$ from equation 3 into equation 4, we get, $V=\frac{q\times E\times d}{q}$
which is $V=E\times d$ ...(5) which is the relationship between electric field intensity and potential difference between two points separated at a distance $d$.