Question

A point charge of +3.0 μC is 10 cm apart from a second point charge of –1.5 μC. Find the magnitude and direction of force on each charge.

Answer 1

Answer:

Magnitude of force on each charge = 4.05 N.

Direction of force = Towards each other along the line joining the two charges.

Explanation:

Given:

First charge, $q_1 = +3.0\ \mu C = +3.0\times 10^{-6}\ C$.

Second charge, $q_2 = -1.5\ \mu C= -1.5\times 10^{-6}\ C$.

Distance between the charges, $r=10\ cm = 0.1\ m$.

According to Coulomb's law, the magnitude of electrostatic force between two charges is given by

$F = k\cdot\dfrac{|q_1||q_2|}{r^2}$.

where,

$k$ = Coulomb's constant = $9\times 10^9\ Nm^2/C^2$.

The direction of this force is along the line joninig the two charges.

It is attractive if the two charges are of opposite polarities and repulsive otherwise.

Using this relation, the magnitude of force on first charge due to second is given by,

$F_{21} = k\cdot \dfrac{|q_1||q_2|}{r^2}\\=9\times 10^9\times \dfrac{3.0\times 10^{-6}\times 1.5\times 10^{-6}}{0.1^2}\\=4.05\ N.$

Since, the first charge is positive and second charge is negative therefore the force is attractive and the direction of this force on second charge is towards the second charge along the line joining the two charges.

The magnitude of force on second charge due to first is given by

$F_{12} = k\cdot \dfrac{|q_2||q_1|}{r^2}\\=9\times 10^9\times \dfrac{1.5\times 10^{-6}\times 3.0\times 10^{-6}}{0.1^2}\\=4.05\ N.$

Again, the first charge is positive and second charge is negative therefore the force is attractive and the direction of this force on second charge is towards the first charge along the line joining the two charges.