Answer:
U = 269.4 kJ
Explanation:
The energy required to place the three charges from infinity is given by:
[tex]U=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}[/tex]
In this case, you have that
q1 = q2 = q3 = q = 20uC
r12 = r13 = r23 = r = 2m
k: Coulomb constant = 8.98*10^9 NM^2/C^2
Then, you replace the values of q, r and k in the equation for the energy U:
[tex]U=3k\frac{q^2}{r}\\\\U=3(8.98*10^9Nm^2/C^2)\frac{(20*10^{-6}C)^2}{2m}=269400\ J=269.4\ kJ[/tex]
hence, the required energy is 269.4 kJ
