# You have two charges that exert a Coulomb force of 42 N on each other in vacuum. If you move them...

## Question:

You have two charges that exert a Coulomb force of 42 N on each other in vacuum. If you move them further apart to double the initial distance and put them in a medium with a dielectric constant of 8, what will be the Coulomb force between them?

## Answer and Explanation:

• {eq}q_{1} {/eq} = first charge.
• {eq}q_{2} {/eq} = second charge.
• r = initial separation between the charges.
• R = final separation between the charges = 2r.
• {eq}k_{c} {/eq} = Coulomb's constant.
• F = initial Coulomb force = 42 N.
• F' = final Coulomb force.
• k = dielectric constant of medium = 8.
• In the presence of dielectric medium, the magnitude of Coulomb force decreases by a factor of k from its original value of force in vacuum.;

According to Coulomb's Law,

{eq}\begin{align} F &= \dfrac{k_{c}q_{1}q_{2}}{r^{2}}.\\ F' &= \dfrac{k_{c}q_{1}q_{2}}{kR^{2}}.\\ &= \dfrac{k_{c}q_{1}q_{2}}{k(2r)^{2}}.\\ &= \dfrac{k_{c}q_{1}q_{2}}{4kr^{2}}.\\ &= \dfrac{F}{4k}.\\ &= \dfrac{42}{4\times 8}.\\ &= 1.31\ N \end{align} {/eq}