Question

A physicist found that a force of 0.69 N was measured between twocharged spheres. The distance between the spheres was 1.3 m. Calculatethe force between the two spheres when the distance was decreased to75 cm.a. 1.2 Nb. 1.6 Nc. 3.3 Nd. 2.1 N

Answer 1

Given that the force is

$F_1=\text{ 0.69 N}$

The distance between the charged spheres is

$r_1=\text{ 1.3 m}$

We have to find the force when the distance is

$\begin{gathered} r_2=\text{ 75 }cm\text{ } \\ =0.75\text{ m} \end{gathered}$

The force is given by the formula

$F=\frac{Kq1q2}{r^2}$

Here K is Coulomb's constant.

q1 and q2 are the charges of the charged sphere.

Also,

$\begin{gathered} kq1q2=Fr^2 \\ =F_1(r_1)^2 \\ =0.69\times(1.3)^2 \\ =1.1661Nm^2 \end{gathered}$

The force when distance is r2 will be

$\begin{gathered} F_2=\frac{kq1q2}{(r_2)^2} \\ =\text{ }\frac{1.1661}{(0.75)^2} \\ =2.1\text{ N} \end{gathered}$

Thus, option d is correct.