There is an electric field $ E $ in x – direction. If the work done on moving a charge of $ 0.2C $ through a distance of 2 m along a line making an angle $ 60^\circ $ with x – axis is 4 J. Then, what is the value of $ E $
(A) $ \sqrt 3 N/C $
(B) 4 N/C
(C) 5 N/C
(D) 20 N/C

Hint : Work done by a body considers only the displacement moved in the direction of the force. The force acting on a positive charge due to an electric field is parallel and in the direction of the electric field.

Formula used: In this solution we will be using the following formula;
 $ F = qE $ where $ F $ is the force acting on a charge in an electric field, $ q $ is the value of the charge, and $ E $ is the electric field at the location of the charge.
 $ W = Fd $ where $ W $ is the work done by a force on an object, $ F $ is the force acting on the object, and $ d $ is the distance moved by the object while the force acts on it in the direction of the force.

Complete step by step answer:
The work done due on a charge to an electric field along x axis, after the charge moved 2m along a line making angle 60 degree to the x –axis is 4 J, the value of the electric field is to be determined.
We must note the force due to an electric field acts along the line of the electric field.
 $ W = Fd $ where $ W $ is the work done by a force on an object, $ F $ is the force acting on the object, and $ d $ is the distance moved by the object while the force acts on it in the direction of the force.
But $ F = qE $ where $ q $ is the value of the charge, and $ E $ is the electric field at the location of the charge.
Hence,
 $ W = qEd $
However since $ d $ must be along the x axis, then
 $ d = 2\cos 60^\circ = 1m $
Hence, by inserting all known values into above equation, we have
 $ 4 = 0.2E\left( 1 \right) $
 $ \Rightarrow E = \dfrac{4}{{0.2}} = 20N/C $
Thus, the correct option is D.

Note:
For clarity, we used $ d = 2\cos 60^\circ $ because the direction of the charge was at an angle 60 degrees to the horizontal (x axis), which was the direction of the force. Hence, the component of the distance in the direction of the force must be used.