Question

a) Calculate the potential energy of a charge +6nC placed at 0.07m from a charge +76 x10^-6C. If the same charge if at 0.07m from a charge -88 x10^-6C, find its potential energy.

b) Find the distance from a charge -8 x10^-6C to a charge -4 x10^-9C if there is a potential energy of 9.5kJ between them. Find the initial force on charge -4 x10^-9C.

Answer 1

a) The potential energy of two point charges is given as follows:

$$W = k\dfrac{q_1q_2}{r}$$

where $q_1,q_2$ are the charges, $r$ is the distance between them, $k = 9\times 10^9 Nm^2/C^2$ is the Coulomb constant. Substituting the numbers from the task, obtain:

1.

$$W =9\times 10^9\cdot \dfrac{6\times10^{-9}\cdot 76\times10^{-6}}{0.07} \approx 0.06J$$

2.

$$W =9\times 10^9\cdot \dfrac{6\times10^{-9}\cdot (-88)\times10^{-6}}{0.07} \approx -0.07J$$

b) Expressing the distance from the first formula, find:

$$r = \dfrac{kq_1q_2}{W}$$

Substituting the numbers, get:

$$r = \dfrac{9\times 10^9\cdot (-8)\times 10^{-6}\cdot (-4)\times 10^{-9}}{9.8\times 10^3} \approx 2.9\times 10^{-8}m$$

The force is given by the Coulomb's law:

$$F = k\dfrac{q_1q_2}{r^2} = \dfrac{W}{r} = \dfrac{W^2}{kq_1q_2}\\ F = \dfrac{(9.8\times 10^3)^2}{9\times 10^9\cdot (-8)\times 10^{-6}\cdot (-4)\times 10^{-9}} \approx 3.3\times 10^{11}N$$

Answer. a) 0.06J and -0.07J. b) $2.9\times 10^{-8}m$ and $3.3\times 10^{11}N$.