Question

Two charges of `7 C` and `5 C` are positioned on a line at points `2` and `-4`, respectively. What is the net force on a charge of `-4 C` at `0`?

Answer 1

`=51.75*10^9N`

Explanation:

By Coulomb's law we Know that force between two charges Q and q situated at a distance r is given by the formula `F=k(qQ)/r^2`,where k =Coulomb's constant=`9*10^9Nm^2C^-2`

Applying this formula the attractive force of 7C on -4C
=`9*10^9*4*7/2^2N=63*10^9N`
The attractive force of 5C on -4C =`9*10^9*4*5/4^2N=11.25*10^9N`

These two forces are oppositely directed so the net force will be towards right and its magnitude `=(63-11.25)*10^9N=51.75*10^9N`