Question

charge q1 and q2 are 0.40 m apart. suppose q1 is -5.00 nC and q2 is -2.00 nC. Find the equilibrium for a positive charge placed between them

Answer 1

0.245 m

Step-by-step explanation:

We are given that

`q_1=-5nC=-5* 10^(-9) C`

Where
`1nC=10^(-9) C`

`q_2=-2.00nC=-2* 10^(-9)C`

Distance between q1 and q2=0.40 m

We have to find the equilibrium for a positive charge placed between them.

Let q be the positive charge placed at distance x from q1 .

r1=x

r2=0.40-x

At equilibrium

Total force =0

`(kq* 5* 10^(-9))/(x^2)-(kq* 2* 10^(-9))/((0.40-x)^2)=0`

Where electric force=
`(kq_1q_2)/(r^2)`

`(5* 10^(-9)* kq)/(x^2)=(2* 10^(-9)* kq)/((0.40-x)^2)`

`5(0.40-x)^2=2x^2`

`0.8-4x+5x^2=2x^2`

`5x^2-2x^2-4x+0.8=0`

`3x^2-4x+0.8=0`

`x=(4\pm√(16-4* 3* 0.8))/(3(2))`

By using the formula

`x=(-b\pm√(b^2-4ac))/(2a)`

`x=(4\pm 2.5298)/(6)`

By solving we get

`x=1.0883,0.245`

`x=1.0883>0.40`

It is not possible

Therefore, the equilibrium point for a positive charge placed between them is given by

x=0.245 m