In this problem, you will be asked to use the given diagram(Figure 1) to calcula

Question

In this problem, you will be asked to use the given diagram(Figure 1) to calculate the work done by the electric field E  on a particle of charge q and see for yourself whether that work appears to be trajectory-independent. Recall that the force acting on a charged particle in an electric field is given by F =E q.

Recall that the work W done on an object by a constant force is

W=Fdcos,

where F is the magnitude of the force acting on the object, d is the magnitude of the displacement that the object undergoes, and is the angle between the vectors F  and d .

Consider a uniform electric field E  and a rectangle ABCD, as shown in the figure. Sides AB and CD are parallel to E  and have length L; let be angle BAC.

Now assume that the particle "chooses" a different way of traveling. Calculate the total amount of work WADC done by the electrostatic force on the charged particle as it moves from A to D to C.

Express your answer in terms of some or all the variables E, q, L and .

Explanation / Answer

I will assume that the rectangle has AB and CD as the lengths, i.e., L and BC and AD as the breadths, say, b

Now, it needs to be understood that the relation for the work done on a particle by a given force F over a displacement of s is given by the dot product of the force vector and the displacement vector.

Hence, W = Fdcos

As the charged particle is travelling from A to D to C, we will take the parts of the journey separately and determine the work done.

For A to D, here the direction of the displacement would be along AD will the direction of the electric field is along AB as has been given.

That is, the force acting on the particle and the displacement happening would be perpendicular to each other.

Hence. dW = Eq b cos90 = 0

For the D to C,

The direction of the electric field is along AB or DC and the displacement in this case is also happening along the same direction. That is, the two vectors are making an angle of 0 degrees with each other.

Hence W = EqL cos0 = EqL

NOTE: No matter what the path is chosen, the magnitude of the work done would remain same. If the particle travels along the diagonal, then the vectors would be at an angle of to each other.

Hence W = Eq D cos where D is the diagonal lenght. If we subsitute the value of cos = L /D, we would obtain the same relation of EqL

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