Question

Objects 1 and 2 attract each other with an electrostatic force of 8 units. If the charge of Object 1 is multiplied by 1 AND the charge of object 2 is multiplied by 3 AND the distance separating Objects 1 and 2 is divided by 4, then the new electrostatic force will be _____ units.

Answer 1

The new electrostatic force will be 384 units.  

The electrostatic force of attraction between two charges $q_{1}$(charge of object 1)  and $q_{2}$(charge of object 2) separated by a distance d is given by

$F = \frac{q_{1} q_{2}}{4\pi E_{0} d^{2} }$

$E_{0} = 8.854 * 10^{-12} C^{2} N^{-1} m^{-1}$ is the permittivity of free space.

Initially, we have,

$\frac{q_{1} q_{2}}{4\pi E_{0} d^{2} } =F_{1} = 8 units$  

Now, if the charge of object 1 is multiplied by 1, the charge of object 2 is multiplied by 3, and the distance separating objects 1 and 2 is divided by 4, we have, $q_{1} =q_{1}, q_{2} =3q_{2} , d=\frac{d}{4}$.

The new electrostatic force will be,

$F_{2} = \frac{q_{1} 3q_{2}}{4\pi E_{0} (\frac{d}{4} )^{2} }$ units.

We have, $F_{1} = 8 units= \frac{q_{1} q_{2}}{4\pi E_{0} d^{2} }$

Hence,

$F_{2} =\frac{8*3}{(\frac{1}{4}) ^{2} } units= 384 units.$

Hence, the new electrostatic force will be 384 units.

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