Question

Two identical charges are placed at the 2 corners of an equilateral triangle. the potential energy of the system is u. the work done in bringing an identical charge from infinite to the third vertex is

Answer 1

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