Question

Two small, identical particles have charges Q1 = +5.5 μC and Q2 = -17μC. The particles are conducting and are brought together so that they touch. Charge then moves between the two particles so as to make the excess charge on the two particles equal. If the particles are then separated by a distance of 60 mm, what is the magnitude of the electric force between them?

Answer 1

Step-by-step explanation:

Given

Charge
`Q_1=+5.5\mu C`

`Q_2=-17\ mu C`

When the two charges are touch then the net charge on the two sphere will be same after some time

`Q=(Q_1+Q_2)/(2)`

`Q=(5.5-17)/(2)=-5.75\ \mu C`

Now the force between this tow charges are when they are separated with
`d=60\ mm`

`F=(kq_1q_2)/(r^2)`

`F=(9* 10^9* (-5.75)^2)/((60* 10^(-3))^2)`

`F=82.65\ N`