Question

A point charge `q_1 = 4.00 nC` is placed at the origin, and a second point charge `q_2 = -3.00 nC`, placed on the x-axis at x = + 20.0 cm. A third point charge `q_3 = 2.00 nC` is placed on the X-axis between `q_1, and q_2`. (Take as zero the potential energy of the three charges when they are infinitely far apart).
(a) What is the potential energy of the system of the three charges if `q_3` is placed at x= + 10.0 cm?
(b) Where should `q_3` be placed to make the potential energy of the system equal to zero?

Answer 1

Correct Answer - B::C::D
a. `U=k((q_1q_2)/r_12+(q_1q_3)/r_13+(q_2q_3)/r_23)`…i
Where `k=1/(4piepsilon_0)`
b. Suppose `q_3` is placed at coordinate `x( gt 0.2m or 20cm)` them in eqn i of part a put
`U=0, r_12=0.2m, r_13=x`
and `r_23=(x-0.2)`
Now, solving eq i we get the desired value of x.