Question

A uniform electric field exists between two charged plates as shown in the figure. What should be the work done in moving a charge q along the closed rectangular path ABCDA.

Answer 1

Hint: Work done is a property that depends on displacement of the object the force is subjected on. $ work done = force \times displacement $ . We need to find the displacement the charge travels through the loop ABCDA.

Formula Used: We will be using the notion that force is conservative and work done by an electric charge $ q $ in an uniform electric field $ E $ is given by $ w = Ed\cos \theta $ where $ w $ is the work done by the charge, $ d $ is the displacement due to the force applied on the charge (which in this case is an electrostatic force), and $ \theta $ is the angle between the two vectors.

Complete Step by Step solution
We know, in a uniform electric field the field lines are going to move from positively charged plate to the negatively charged one. Now suppose you are placing a charge in between the plates like shown in the figure below. Also consider an imaginary rectangular path ABCD with dimensions, $ AB = d_1 = CD $ and $ BC = d_2 = DA $ .

Now let us consider the work done to move the charge from point A to B between the plates.
 $ w = Ed_1\cos 0^\circ $
Here $ \theta = 0^\circ $ , because the work done on the point to move from A to B and the electric field lines are in the same direction.
We also know that, $ \cos 0^\circ = 1 $ . So, $ w = Ed_1(1) $
 $ w = Ed_1 $
Similarly, along the path $ BC = d_2 $ , the angle between the two vectors is, $ \theta = 90^\circ $ . $ w = Ed_2\cos 90^\circ $
 $ w = Ed_2(0) $
And thus, the work done to move charge on the path BC,
 $ w = 0J $
Similarly for path CD we can see that the angle between the vectors $ \theta $ is $ 180^\circ $ . We know that $ \cos 180^\circ = - 1 $ . Thus, making work done along the path CD, $ w = Ed_1\cos 180^\circ $ .
 $ w = Ed_1( - 1) $
 $ w = - Ed_1 $
Also, for the path DA, the angle $ \theta $ is $ 270^\circ $ . Thus, the work done can be given by $ w = Ed_2\cos 270^\circ $ . Now we can find the work done along the path DA as $ w = Ed_2(0) $ . Since, $ \cos 270^\circ = 0 $ .
 $ w = 0 $
Now that we have the work done to move the charge $ q $ along the closed path ABCDA. Let us calculate the total work done by adding the work done to move the charge along each side.
 $ w = Ed_1 + 0 + ( - Ed_1) + 0 $
 $ w = 0 $
Thus, the work done is zero. No actual work is done to move a charge around the closed path ABCDA as shown in the figure.

Note
Alternate solution-
Since we can see that the charge is supposed to travel along the path ABCDA. The particle starts from A and ends up back in A, thus making the displacement $ d = 0 $ . Work is a quantity whose magnitude depends on the displacement, thus the work done is
 $ w = Ed\cos \theta $
 $ w = E(0)\cos \theta $
Thus, making the work done zero. This simple logic can be used to solve the problem faster.