Question

The unit of electric field is not equivalent to
(A). $N\,{{C}^{-1}}$
(B). $J\,{{C}^{-1}}$
(C). $V{{m}^{-1}}$
(D). $J\,{{C}^{-1}}{{m}^{-1}}$

Answer 1

Every charged particle in an electric field experiences a force on it. Electric field is related to different quantities. It depends on the force, the distance from the centre, work done due to the electric field and the potential difference in an electric field. Using these relations, we determine the different units of the electric field in terms of other units.

Formulas used:
$E=\dfrac{V}{d}$
$E=\dfrac{F}{q}$
$W=qV$

Complete answer:
Electric field is defined as the work done to move a unit charge from infinity to a point in the field. It is a vector quantity and its direction is the direction in which positive charge moves in a field. As electric fields can be calculated using different relations, it can have different units.
Electric field can be calculated as-
$E=\dfrac{V}{d}$ - (1)
Here, $E$ is the electric field
$V$ is the potential difference between two points
$d$ is the distance between the two points between which potential is calculated
From the above equation, the unit of potential is volt ($V$) and the unit of distance is metres ($m$). Therefore, the unit of electric field will be $\dfrac{V}{m}=V{{m}^{-1}}$.
Electric field can also be calculated as-
$E=\dfrac{F}{q}$
Here, $F$ is the electric force acting on a charged body in the electric field
$q$ is the charge on the charged body
From the above equation, the unit of force is newton ($N$) and the unit of charge is coulomb ($C$). Therefore, the unit of electric field will be $\dfrac{N}{C}=N{{C}^{-1}}$.
We know that,
$W=qV$
Here, $W$ is the work done
Substituting $V$ from eq (1), we get,
$\begin{aligned} & W=qEd \\\ & \Rightarrow E=\dfrac{W}{qd} \\\ \end{aligned}$
From the above equation, the unit of work is joules ($J$), the unit of charge is coulomb ($C$) and the unit of distance is metres ($m$). Therefore, the unit of electric field will be $\dfrac{J}{Cm}=J\,{{C}^{-1}}{{m}^{-1}}$
Therefore, the unit which is not a unit of electric field is $J\,{{C}^{-1}}$.

Hence, the correct option is (B).

Note:
Electric field is different for different types of conductors. It is represented by electric lines of forces which are open curves coming out of the positive charge and terminating into negative charge. Tangents drawn to electric lines of forces give us the direction of the electric field.