Question

SI unit of the electric field is …………………

& (a)~N{{C}^{-3}} \\\ & (~b)~N{{C}^{-2}} \\\ & (c)~N{{C}^{-1}} \\\ & (d)\text{ }NC~ \\\ \end{aligned}$$

Answer 1

Hint: Electric field is defined as force per unit positive charge. Electric lines of force visually describe the electric field and it is denoted by E.

Complete step by step answer:
If a force of electric origin is utilized on a motionless charged body placed at that point, the electric field is said to remain at a point. The electrostatic force created between two charged frames alike without any explicit contact between them. The nature of action at distance force can be understood by announcing the concept of electric field.

The electric intensity $\overset{\to }{\mathop{E}}\,$ at a point is explained as the force accomplished by a unit positive charge placed at that point, without distressing the situation of origin charge.

Let a test charge ${{q}_{0}}$ experiences a force $\overset{\to }{\mathop{F}}\,$ at that point P. Then the electric field at that point is given as –

$\overset{\to }{\mathop{E}}\,\text{ = }\dfrac{\overset{\to }{\mathop{F}}\,}{{{\text{q}}_{\text{0}}}}\text{ }......\text{ (1)}$

In the equation (1), the test charge ${{q}_{0}}$ may disturb the charge distribution of the source charge and hence charge the electric field $\overset{\to }{\mathop{E}}\,$ which we want to measure. The test charge ${{q}_{0}}$ must be small enough so that it doesn’t change the value of $\overset{\to }{\mathop{E}}\,$.

The SI unit of electric field $\overset{\to }{\mathop{E}}\,\text{ = }\dfrac{\text{Force}}{\text{Charge}}\text{ = }\dfrac{\text{N}}{\text{C}}\text{ = N}{{\text{C}}^{\text{-1}}}$

So the correct option is (c).

Additional Information:
The SI unit of electric field $\left( \text{N}{{\text{C}}^{\text{-1}}} \right)$ is also equivalent to volt per metre $\left( \text{V}{{\text{m}}^{\text{-1}}} \right)$.

The dimensions of electric field $\overset{\to }{\mathop{E}}\,$ can be termed as –

$\begin{aligned} & \left[ \text{E} \right]\text{ = }\left[ \dfrac{\text{ML}{{\text{T}}^{\text{-2}}}}{\text{C}} \right]\text{ }......\text{ (2)} \\\ & \left[ \text{E} \right]\text{ = }\dfrac{\text{ML}{{\text{T}}^{\text{-2}}}}{\text{A}\text{.T}}\text{ = }\left[ \text{ML}{{\text{T}}^{\text{-3}}}{{\text{A}}^{\text{-1}}} \right] \\\ \end{aligned}$

Note: Always remember the electric field depends upon the force and the charge of the body and note that the charge which you take to find the value of electric field should be very small.Student should remember that SI unit of intensity of electric field and SI unit of electric field are some, don’t get confused.