Question

If you carry out the integral of the electric field S for a closedpath like that shown in Fig. Q23.9, the integral will alwaysbe equal to zero, independent of the shape of the path and independent of where charges may be located relative to the path. Explain why

Short answer

The finite integration over a closed path with the initial value and the final value is equal, then the value of the integration is equal to zero.

Step-by-step solution

About electric field 

electric field, an electric property associated with each point in space when charge is present in any form.The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Determine why the integral will always be equal to zero or independent of shape of path

As we know the electric field is a conservative filed- Also, We know that the electric field is given by the following relation:

So, we solve for the given integral on the closed path:

As we know the closed integral on the gradient of the electric potential is equal to zero. Hence, the potential initial potential

is equal to the ?nal potentiaL We know that the finite integration over a closed path with the initial value and the final value is

equal, then the value of the integration is equal to zero. 80, the work done is independent on the path and is zero for the

closed paths-

Result

As We know the closed integral on the gradient of the electric potential is equal to zero- Hence, the potential initial potential is equal to the final potential- We know that the finite integration over a closed path with the initial value and the final value is equal, then the value of the integration is equal to zero.

So, the Work done is independent on the path and is zero for the closed paths.