Question

Three +0.12C charges from an equilateral triangle 1.7m on a side. Using energy supplied

At the rate of 0.83KW, how many days would be required to move one of the charges to

the midpoint of the line joining the other two charges?

Answer 1

The potential at the apex due to the other two charges is$k\times \frac{q1}{r} + k\times \frac{q2}{r} = 9.0\times 10^9 \times(\frac{0.12}{1.70} + \frac{0.12}{1.70}) = 1.27\times10^9 V$

The potential between them is $9.0\times10^9 \times(\frac{0.12}{0.85} + \frac{0.12}{0.85}) = 2.54\times10^9 V$

so the potential difference is $2.54\times10^9 - 1.27\times10 9 = 1.27\times10^9 V$

so the energy $= V\times q = 1.27\times10^9 \times0.12 = 1.52\times10^8 J$

Since P =$\frac{Energy}{t}$ then t $= \frac{E}{P} = \frac{1.52\times10^8}{0.83\times10^3} = 1.831\times10^5 s$

converting to days (1 day = 86400s) we get t =$\frac{ 1.831\times10^5}{86400} = 2.11$ days