Question

Three charges q = 0.140 C form an equilateral triangle, 1.50 m on a side. Using energy that is supplied at the rate of 0.64 kW, how many days would be required to move one of the charges to the midpoint of the line joining the other two charges? (Do not enter a unit)

Answer 1

The potential at the apex due to the other two charges is k*q1/r + k*q2/r

= 9.0x10^9*(0.14/1.50 + 0.14/1.50) = 1.68x10^9V

The potential between them is 9.0x10^9*(0.14/0.75 + 0.14/0.75) = 3.36x10^9V

so the potential difference is 3.36x10^9 - 1.68x10^9 = 1.68x10^9V

so the energy = V*q = 1.68x10^9*0.14 = 2.352x10^8J

Since P = Energy/t then t = E/P = 2.352x10^8/0.64x10^3 = 3.675x10^5s

converting to days (1 day = 86400s) we get t = 3.675x10^5/86400 = 4.253 days