Answer:
the final potential energy of this system is 3U0/10
Explanation:
We are given
charge at left end and another test charge at point p
Potential energy is given by = [tex]\frac{k*Q1*Q2 }{R}[/tex]
where k is electrostatics constant = [tex]9 *10^9[/tex]
Q1 = first charge , Q2= test charge
R= distance between charges
potential at point p
U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1
now the test charge moves to point R
using Pytahgoreou theorem
R(distance) = [tex]\sqrt{8^2 + 6^2}[/tex] = 10
New Potential energy
U1 = kq1*q2 / 10
substituting kq1q2 = 3U0 from 1
U1 = 3U0/10
So this is the final potential energy of this system.
