Question

A -2.37 µC point charge and a 4.74 µC point charge are a distance L apart.

Where should a third point charge be placed so that the electric force on that third charge is zero? (Hint: Solve this problem by first placing the -2.37 µC point charge at the origin and place the 4.74 µC point charge at x = −L.)

Answer 1

`x=2L\pm \sqrt{((-2L)^2-4* 1*(-L^2))/(2* 1) }`

Step-by-step explanation:

Given:

• charge on first particle,
  `q_1=-2.37* 10^(-6)\ C`

• charge on the second particle,
  `q_2=4.74* 10^(-6)\ C`

• distance between the two charges = L

Now the third charge must be placed on the line joining the two charges at a distance where the intensity of electric field is same for both the charges that point will not lie between the two charges because they are opposite in nature.

`E_1=E_2`

`(1)/(4\pi\epsilon_0) * (q_1)/(x^2) =(1)/(4\pi\epsilon_0) * (q_2)/((L+x)^2)`

`(2.37)/(x^2) =(4.74)/(L^2+x^2+2xL)`

`2x^2=L^2+x^2+2xL`

`x^2-2L.x-L^2=0`

`x=2L\pm \sqrt{((-2L)^2-4* 1*(-L^2))/(2* 1) }`