Two spheres of electric charges +2 nC and -8 nC are placed at a distance d apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

Options

A. `(4d)/(3)`
B. `(3d)/(4)`
C. d
D. `(d)/(2)`

Answer

B

Explanation

In the first condition,
Given, `q_(1)=+2nC=2xx10^(-9)C`
`q_(2)=-8nC`
`=-8xx10^(-9)C`
`F=(kq_(1)q_(2))/(r^(2))`
`F=(k(2xx10^(-9))xx(8xx10^(-9)))/(a^(2))`
`F=(k(16xx10^(-18)))/(a^(2))`
In the second condition,
`F_(1)=(kq_(1)q_(2))/(d^(2))`
After touching of sphere to each other the total charge `=2-8=-6nC`
Charge on each sphere = 3nC
`= 3xx10^(-9)C`
`F_(1)=(k(3xx10^(-9))(3xx10^(-9)))/((d^(1))^(2))`
`F_(1)=(k(9xx10^(-18)))/((d^(1))^(2))`
From Eqs.(i) and (ii), we get
`(k(16xx10^(-18)))/(d^(2))=(k(9xx10^(-18)))/((d^(1))^(2))`
`(d^(2))/(d^(1^(2)))=(16xx10^(-18))/(9xx10^(-19))`
`(d^(2))/(d^(1^(2)))=(16)/(9)`
`(d)/(d^(1))=sqrt((16)/(9))`
`(d)/(d^(1))=(4)/(3)`
`d=(4),/(3)d^(1)`
`d^(1)=(3)/(4)d`