Question

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newtons do they experience when their separation is 0.7 m?

Answer 1

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude $q_{1},q_{2}$ the magnitude of force between them is given by

$F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}$

where

$k_{e}$ is constant

$r$ is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

$F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208$

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

$F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}$

Applying value of the constant we get

$F_{1}=\frac{62.208}{0.7^{2}}$

Thus $F_{2}=126.955Newtons$