Question

Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?

Answer 1

x = 0.006 m

Step-by-step explanation:

The potential at one point is given by

V = k ∑
`q_(i) / r_(i)`

remember that the potential is to scale, let's apply to our case

V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)

in this case they indicate that the potential is zero

0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + ​​3 10⁻⁶ / x)

3 / x = + 2 / 10⁻² + ​​3 / 10⁻²

3 / x = 500

x = 3/500

x = 0.006 m