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# 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.

Last updated date: 24th Jun 2024
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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.