Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +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? =

  x = -3 cm

Explanation:

The electrical potential is the sum of the potentials of each charge

       V = k ∑

let's apply this to our case where the potential is V = 0 for x = 0

         0 = k (q₁ / (x₁-0) + q₂ / (x₂-0) + q₃ / (x₃-0))

in our case

q₁ = + 2.0 10⁻⁶ C

q₂ = - 6.0 10⁻⁶ C

q₃ = + 3.0 10⁻⁶ C

x₁ = -1.0 cm = 1.0 10⁻² m

x₂ = +2.0 cm = 2.0 10⁻² m

we substitute in the equation

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

          3 10⁻⁶ / x = 2 10⁻⁴ - 3 10⁻⁴

          3 10⁻⁶ / x = -1 10⁻⁴

           x = - 3 10⁻² m

           x = -3 cm

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

the answer is a)

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