Question

Two charges, are placed on the x-axis (see below). If charge B is moved to `x=oo` then, the potential at the origin is `200 "kV"`. If charge B is moved to `x = 7` m, the potential at the origin is equal to `0"V"`. What is the charge of B?

I'm not sure what it's asking

Answer 1

`q_B =-1.56times 10^-7" C"`

Explanation:

When the charge B moves to `x = oo` it does not contribute to the potential at the origin. Thus, contribution of charge A to the potential at the origin is 200 V.

When the charge B is at `x = 7" m"`, the potential at the origin is zero. This means that the potential at the origin due to charge B now is -200 V. So

`1/{4pi epsilon_0} q_B/{7 " m"} = -200" V"`

so that

`q_B = 4pi epsilon_0 times (-200 " V")times 7 " m" = -{1400" V-m"}/{0.899times 10^10" V-m/C"}= -1.56times 10^-7" C"`