Question

The unit of electric field strength is
(A) newton/coulomb
(B) newton/ampere
(C) volt/coulomb
(D) joule/coulomb

Answer 1

Hint
The electric field strength or the electric field intensity at a point is given by the ratio of the force due to the electric field and the charge that is placed at that point. So we can take the ratio of the units of force and charge to find the answer.
In this solution we will be using the following formula,
$\Rightarrow E = \dfrac{F}{q}$
where $E$ is the electric field,
$F$ is the force and $q$ is the charge.

Complete step by step answer
The electric field strength is a vector quantity as it has both the magnitude and direction. Let us consider a charge $Q$ is placed which creates an electric field in its surroundings. Now if a test charge is brought from infinity to that point then that charge will experience a force due to the presence of the field created by the charge $Q$. This field is called the electric field of the charge $Q$. It is denoted by the letter $E$.
The magnitude of this charge is given by the force that is experienced by the test charge divided by the value of the test charge. That is, the magnitude of the electric field strength is, $\Rightarrow E = \dfrac{F}{q}$.
The unit of force is given by newton and the unit of charge is given by coulomb.
Now in this formula if we substitute the unit of force and the unit of charge, we get the unit of electric field strength as, $E = \dfrac{{newton}}{{coulomb}}$
Therefore the unit of electric field strength is newton/coulomb.
So the correct answer is option (A).

Note
 The force between the test charge and the charge $Q$ is given by the formula,
$F = \dfrac{{kQq}}{{{R^2}}}$ where $R$ is the distance between the charge. So to find the electric field strength we divide the force with the test charge and get,
$E = \dfrac{1}{q}\dfrac{{kQq}}{{{R^2}}} = \dfrac{{kQ}}{{{R^2}}}$
So the $q$ gets cancelled and hence the electric field strength is independent of the value of the test charge.