Question

Two point charges of equal and opposite charge ±10−8 C sit a distance of 20 cm apart. What is the strength of the force they exert on each other? If one of the point charges is sitting on a table and has a mass of .010 kg, how close must the second charge be to lift the charge up off the table? What is the electric field strength required to lift the charge of the table?

I got the strength of force to be equal to 2.25x10^-5 N, but am unsure how to continue this question.

Answer 1

Given that,

q1 = + 10^-8 C, q2 = -10^-8 C

Distance between them, r = 20 cm = 0.20 m

According to Coulmob's law, electrostatic force between the two charges is given as -

F = k*q1*q2 / r^2

= (9.0 x 10^9 x 10^-8 x (-10^-8)) / 0.20^2 = -2.25 x 10^-5 N

Negative sign shows that there is a force of attraction between the charges.

Now, mass of the charge sitting on the table, m = 0.010 kg

To lift this mass from the table, we have to balance its weight which is equal to 'm*g'.

Suppose the distance between the charges in this case = d

So we have -

k*q1*q2 / d^2 = m*g

=> (9.0 x 10^9 x 10^-8 x 10^-8) / d^2 = 0.010*9.8

=> d^2 = (9.0 x 10^9 x 10^-8 x 10^-8) / (0.010*9.8) = 91.84 x 10^-7 = 9.184 x 10^-6

=> d = 3.03 x 10^-3 m = 3.03 mm (Answer)

Again, suppose E is the magnitude of the electric field.

So, force on charge q due to this electric field = E*q

Equalize this with the weight of the charge -

E*q = m*g

=> E = (m*g) / q = (0.010*9.8) / (10^-8) = 9.8 x 10^6 N/C (Answer)