Question

1. A point charge q1 = +5.60 μC is at origin. How far should the second point charge of +8.10 μC be placed to have electric

potential energy of 0.900 J?

2. A point charge has a charge of 4.00 x 10-11 C. At what distance from the point charge is the electrical potential

(a) 8.0V?

(b) 16.0V?

Answer 1

1. The potential energy of electrostatic interaction

$$E_p=k\frac{q_1q_2}{r}$$

Thus, the distance between charges

$$r=k\frac{q_1q_2}{E_p}\\ =9*10^9*\frac{5.60*10^{-6}*8.10*10^{-6}}{0.900}=0.45\:\rm m$$

2. The electric potential

$$V=k\frac{q}{r}$$

(a)

$$r=k\frac{q}{V}=9*10^9*\frac{4.00*10^{-11}}{8.0}=0.045\: \rm m$$

(b)

$$r=k\frac{q}{V}=9*10^9*\frac{4.00*10^{-11}}{16.0}=0.0225\: \rm m$$