Question

There is an electric field $E$ in the + $x$ -direction. If the work done by the electric field in moving a charge $0.2{\text{ }}C$ through a distance of 2m along a line making an angle of $60^\circ$ with the x-axis is 1J, what is the value of E in $N/C$ ?
A) $\sqrt 3 \,N/C$
B) $4\,N/C$
C) $5\,N/C$
D) None of these

Answer 1

Hint : Work is required to be done when moving a charge in an electric field. Work is only done when moving a charge in an electric field in a direction parallel to the electric field.

Formula used: In this solution, we will use the following formula,
Work done in moving a charge in an electric field: $W = q\Delta V$ where $q$ is the magnitude of the charge and $\Delta V$ is the potential difference between the final and initial points.

Complete step by step answer
We’ve been given that an electric field exists in the + $x$ -direction and that the work done by the electric field in moving a charge of magnitude $0.2{\text{ }}C$ through a distance of 2m along a line making an angle of $60^\circ$ with the $x$ -axis is 1J. We know the work done in moving a charge in an electric field is calculated as
$\Rightarrow W = q\Delta V$
Since we only want to calculate the distance moved by the charge in the direction of the electric field, we can calculate the potential difference as the dot product of the electric field and the distance moved by the charge:
$\Rightarrow \Delta V = E.r$
$\Rightarrow \Delta V = Er\cos \theta$ where $\theta$ is the angle formed by the electric field and the line on which the charge is moving.
Since we’re moving the charge on a line making an angle of $60^\circ$ , we can substitute $q = 0.2,\,\theta = 60^\circ ,\,r = 2m$ and calculate the work done as
$\Rightarrow W = qEr\cos \theta$
$\Rightarrow 1 = 0.2 \times E \times 2 \times \cos (60^\circ )$
Since $\cos (60^\circ ) = 1/2$ on solving for $E$ , we get
$\Rightarrow E = 5\,N/C$ which corresponds to option (C).

Note
Here we can only calculate the potential difference between the final and the initial point in the way we did if the electric field is constant in space and not varying. We must also take the dot product of the electric field and the distance travelled by the charge when calculating the potential difference since we only take the component travelled by the charge in the direction of the electric field which is in the + $x$ direction.