The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

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The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

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Answer 1 · 2016-01-07T16:53:17

Let $q_{1}$ $q_1$ and $q_{2}$ $q_2$ be the charges, $K$ $K$ be the Coulomb's constant and $r$ $r$ be the distance between the charges when the force is $3.4 N$ $3.4N$ and let $F$ $F$ denote this force.

$F = \frac{k q_{1} q_{2}}{r^{2}}$ $F=(kq_1q_2)/r^2$

$\Rightarrow 3.4 = \frac{k q_{1} q_{2}}{r^{2}}$ $implies 3.4=(kq_1q_2)/r^2$

Now, every other condition remains same except the distance between the charges. The distance between the charges has been increased $4$ $4$ times its original value. Let $r'$ be then new distance between the charges, then
$r'=4r$
Let the new force between the charges be denoted by $F'$ .
$F'=(kq_1q_2)/r_'^2$

$implies F'=(kq_1q_2)/(4r)^2$

$implies F'=(kq_1q_2)/(16r^2)$

$implies F'=(1/16)((kq_1q_2)/r^2)$

Since $3.4=(kq_1q_2)/r^2$

$implies F'=(1/16)(3.4)$

$implies F'=0.2125N$

Hence the force between the two charges becomes $0.2125N$

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The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

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