Question

Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction be when the spheres are moved to 10 cm apart?

Answer 1

Answer:

$3.2\cdot 10^{-8} N$

Explanation:

The inital electrostatic force between the two spheres is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

$F=8\cdot 10^{-9} N$ is the initial force

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

$r'=\frac{r}{2}$

Therefore, the new electrostatic force will be

$F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F$

So the force has increased by a factor 4. By using $F=8\cdot 10^{-9} N$, we find

$F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N$