Question

Calculate the columb force (F1), between two equal charges placed in vacuum separated at a distance square of r meters apart. By what factor does F1 changes when the separation is doubled?

Answer 1

The Coulomb force (F₁) between the two equal charges separated by a distance r from each other in vacuum can be calculated as follows:

$$F_1 = \frac{k q^2}{r^2},$$

where k = 9×10⁹ N·m²·C⁻² is the Coulomb constant.

If the separation between the charges is doubled, i.e.

$$R = 2r$$

then the Coulomb force reduces as follows:

$$F_2 = \frac{k q^2}{R^2} = \frac{k q^2}{4r^2}=\frac{F_1}{4}$$

Answer: reduction by a factor 4.