Question

Consider a coin electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8kC$. Suppose that these equal charges were concentrated in two point charges seperated by
(i) $1cm(\sim \frac{1}{2}\text{diagonal of the one paisa coin})$

(ii) $100m(\sim \text{Length of a long building})$

(iii) ${10}^{6}m\left(\text{Radius of the earth}\right)$

Find the force on each such point charge in this case. What do you conclude from this result?

Answer 1

(i) Given, magnitude of the charge,
$q_1=q_2=34.8kC=3.48\times {10}^{4}C$
Distance between two point charges,
$r=1cm={10}^{-2}m$

So, the force between these point
charges, $F=\frac{k{q}_1{q}_2}{r^2}$

$F=\frac{(9\times {10}^9){(3.48\times {10}^4)}^2}{{\left({10}^{-2}\right)}^2}$

$F=1.09\times {10}^{23}N$

When electrons and protons are separated as point charges, these charges exert an enormous force. It is not easy to disturb electrical neutrality.

Final answer: $1.09\times {10}^{23}N$

(ii) Given, magnitude of the charge,
$q_1=q_2=34.8kC=3.48\times {10}^{4}C$
Distance between two point charges,
$r=100m$
So, the force between these point
charges, $F=\frac{k{q}_1{q}_2}{r^2}$
$F=\frac{(9\times {10}^9){(3.48\times {10}^4)}^2}{{\left(100\right)}^2}$

$F=1.09\times {10}^{15}N$

When electrons and protons are separated as point charges, these charges exert an enormous force. It is not easy to disturb electrical neutrality.

Final answer: $1.09\times {10}^{15}N$

(iii) Given, magnitude of the charge,
$q_1=q_2=34.8kC=3.48\times {10}^{4}C$
Distance between two point charges,
$r={10}^{6}m$
So, the force between these point
charges, $F=\frac{k{q}_1{q}_2}{r^2}$

$F=\frac{(9\times {10}^9){(3.48\times {10}^4)}^2}{{\left({10}^6\right)}^2}$

$F=1.09\times {10}^{7}N$

When electrons and protons are separated as point charges, these charges exert an enormous force. It is not easy to disturb electrical neutrality.

Also, electrostatic force is much higher than gravitational force in general.

Final answer: $1.09\times {10}^{7}N$