Two spheres carrying charges `+ 6 mu C and 9 mu C ` , seperated by a distance d , experience a force of repulsion F . When a charge of ` - 3 mu C ` is given to both the sphere and kept at the same distance as before , the new force of repulsion is

Options

A. 3F
B. `F/9`
C. F
D. `F/3`

Answer

D

Explanation

Let us call spheres as A and B .

According to Coulomb's law , the force repulsion between A and B seperated by a distance d is
`F = 1/(4piepsilon_(0)) ((+CmuC)(+9muC))/(d^(2))" "….(i)`
When a charge `-3 mu C ` is given to both the spheres , then charge on `A 6 mu C - 3 mu C`
` = + 3 mu C`
and on `B = + 9 mu C - 3 mu C = + 6 mu C `
Again by Columb's law , the new force of distance d is
` F' = 1/(4piepsilon_(0)) ((+3muC)(+6muC))/(d^(2)) " " ....(ii)`
Dividing eqn . (ii) by eqn . (i) we get ,
` (F')/F = (1/(4piepsilon_(0)) ((+3muC)(+6muC))/(d^(2)))/(1/(4piepsilon_(0)) ((+6muC)+(9muC))/(d^(2)))`
` = ((+3muC)(+6muC))/((+6muC)(+9muC))= 1/3 `
or F ' = F/3