Question

Charges A and B in the figure are equal. Each charge exerts a force on the other of magnitude . Suppose the charge of B is increased by a factor of 4, buteverything else is unchanged.

Answer 1

Is the question what happens to the forces?

F = kq₁q₂/r², so if q₁ suddenlybecomes 4q₁, the force is multiplied by 4 as well. ByNewton's third law, it applies to both forces; bothare multiplied by 4.

Answer 2

Concepts and reason

The concepts required to solve the given question is electrostatic force.

Initially, write an expression for the electrostatic forces between two charges. Later, increase the magnitude of the second charge four times. Finally, find the magnitude of the force on A.

Fundamentals

The expression for the electrostatic force is as follows:

$F = \frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$

Here, q and q’ are the magnitude of the charges, r is the distance between the charges, and ${\varepsilon _0}$ is the permitivitty of free space.

Assume q and q be the two equal charges placed at point A and B respectively. Assume the separation between the charges be r.

Thus, the force between the two charges using the Coulomb ‘law is obtained by substituting q for q and q’in the equation $F = \frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$ .

$F = \frac{{qq}}{{4\pi {\varepsilon _0}{r^2}}}$

Substitute 4q for q’in the equation $F = \frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$ .

$\begin{array}{c}\\F' = \frac{{q\left( {4q} \right)}}{{4\pi {\varepsilon _0}{r^2}}}\\\\ = \frac{{4qq}}{{4\pi {\varepsilon _0}{r^2}}}\\\\ = 4\left( {\frac{{qq}}{{4\pi {\varepsilon _0}{r^2}}}} \right)\\\\ = 4F\\\end{array}$

Ans:

The magnitude of the force on A is four times the initial force exerted by B.