Question

Define intensity of electric field at any point. Write down its S.I. unit.

Hint:

Electric field intensity relates force and charge; think of it as force per unit charge.

Answer 1

The intensity of the electric field at any point is defined as the force experienced by a unit positive test charge placed at that point. It is a vector quantity that indicates both the magnitude and direction of the force exerted on the charge. Mathematically, the electric field intensity \( \vec{E} \) is given by: \[ \vec{E} = \frac{\vec{F}}{q}, \] where

\( \vec{E} \) is the electric field intensity,
\( \vec{F} \) is the force experienced by the test charge,
\( q \) is the magnitude of the positive test charge placed in the field.
This definition implies that the electric field at a point is independent of the test charge used, provided the charge is small enough not to disturb the existing field.
S.I. Unit:
The S.I. unit of electric field intensity is derived from the units of force and charge. Since force is measured in newtons (N) and charge in coulombs (C), the unit of electric field intensity is: \[ \text{Newton per coulomb (N/C)}. \] Alternatively, electric field can also be expressed in terms of potential difference and distance, giving the unit: \[ \text{Volt per meter (V/m)}. \] Both units are equivalent, as \( 1 \, \text{N/C} = 1 \, \text{V/m} \).
Thus, the electric field quantifies how strongly and in which direction an electric force would act on a positive unit charge placed at any point in space.