A charge of +8.0 mC is moved through a uniform electric field of 20 V/m. The electric field lines are horizontal and point to the left. Calculate the work done along path JKLM, where J to K is 2m to t

The uniform electric field strength is $E = 20 \text{ V m}^{-1}$.
The charge being moved is $q = +8.0 \text{ mC} = +8.0 \times 10^{-3} \text{ C}$.
The electric field lines are horizontal and point to the left.

The work done by a uniform electric field on a charge $q$ is given by $W = q \vec{E} \cdot \vec{d}$, where $\vec{d}$ is the displacement vector. This can also be written as $W = q E d \cos\theta$,