Question

Two charges of -5×10^-9 C and -1.5×10^-9 C are separated by a distance of 43cm. Find the equilibrium position for a third charge of +1.1×10^-8 C by identifying its distance from the first charge q1.

Answer 1

In this question we have given

Charge,
`q_(1)=-5* 10^-9C`

Charge,
`q_(2)=-1.5* 10^-9C`

Charge,
`q_(3) =1.1* 10^-8C`

Distance between Charge
`q_(1)` and
`q_(2)`,d=43cm=.43m

Let the distance between charge
`q_(1)` and
`q_(3)`, is x

Therefore,distance between charge
`q_(2)` and
`q_(3)`, will be=(.43-x)

We know By Couloms law, Force on Charge
`q_(3)` due to charge
`q_(1)` which are seperated by distance d is given by following equation

`F_(1)=(k* q_(1)q_(2))/(d^2)`............(1)

Here, K=
`9* 10^9 Nm^2C^-2`

Therefore,

Force on Charge
`q_(3)` due to charge
`q_(1)`

`F_(1)=-(k* 5* 10^-9C* 1.1* 10^-8C)/(x^2)`..(2)

Similarly,

Force on Charge
`q_(3)` due to charge
`q_(2)`

`F_(2)=-(k* 1.5* 10^-9C* 1.1* 10^-8C)/((.43-x)^2)`..(2)

In equilibrium condition,

`F_(1) =F_(2)`

`-(k* 5* 10^-9C* 1.1* 10^-8C)/(x^2)=-(k* 1.5* 10^-9C* 1.1* 10^-8C)/((.43-x)^2)`

`(5* 10^-9C)/(x^2)=(1.5* 10^-9C)/((.43-x)^2)`

`5* 10^-9C * (.43-x)^2=1.5* 10^-9C* (x)^2`

`3.5x^2-4.3x+.9245=0`

on solving we get

x=.277m

0r x=28cm

The distance between charge
`q_(1)` and
`q_(3)`, is 28cm