There is an electric field $$E$$ in $$x$$-direction. If the work done on moving a charge of $$0.2\,C$$  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 $$E$$ ?

Solution :
Work done in moving the charge, $$W = Fd\,\cos \theta $$
As $$F = qE$$
$$\therefore W = qEd\cos \theta $$
or $$E = \frac{W}{{qd\cos \theta }}$$
Here, $$q = 0.2\,C, d = 2\,m$$
$$\eqalign{ & \theta = {60^ \circ },W = 4\,J \cr & \therefore E = \frac{4}{{0.2 \times 2 \times \cos {{60}^ \circ }}} \cr & = 20\,N/C \cr} $$
Alternative
As we know that potential at any point in the direction of $$\theta $$ and electric field $$E$$ is given by $$dV = - E \cdot dr\,\,\,\left( {{\text{negative sign indicates decreasing potential in direction of electric field}}} \right)$$
So, for the given situation $$dr = d\cos \theta $$
So, $$dV = Ed\cos \theta $$
Now, work done for a charge moving in potential difference $$dV$$ is given by $$W = qdV$$
$$ \Rightarrow W = qEd\cos \theta $$
Given, $$q = 0.2\,C,d = 2\,m,\theta = {60^ \circ },W = 4\,J$$
So, $$4\,J = 0.2 \times E \times 2 \times \cos {60^ \circ }$$
$$ \Rightarrow E = \frac{4}{{0.2 \times 2}} \times 2 = 20\,J$$