Question

Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is u. The work done in bringing an identical charge from infinity to the third vertex is

Answer 1

Answer:

Let 'q' be the magnitude of each charge

Let 'r' be the length of each side of the triangle ABC.

The potential energy of the system of two equal charges placed at a vertices A and B is V.

Hence the work done in bringing an identical charge q from infinity to the third vertex C = work done to overcome the force of repulsion of q placed at A placed at a distance r from it + work done to overcome the force of repulsion of q placed at B at the same distance r from it,

W=V+V=2V

hence the answer is2 volt.

Answer 2

Step-by-step explanation:

Initially

U = kq1q2 /r = kq² /r

When 3 rd charge is brought

U' = kq1q2/r + kq2q3/r + kq1q3 /r

As all charges are same and all side length are same

U' = 3kq²/r = 3U

Tip :- We have to find work done = Change in potential

       W = U' - U = 2U

                         Hope you understand      

This is Habel Sabu ....... Isn't it the Brainliest