Question

A charge of +2.20 μC is at the origin and a charge of -3.20 μC is on the y axis at y = 40.0 cm. 1) What is the potential at point a, which is on the x axis at x = 40.0 cm? (Express your answer to three significant figures.)

Answer 1

Explanation:

Formula to calculate the potential at point "a" is as follows.

      $V_{a} = \frac{kq_{1}}{r_{1}} + \frac{kq_{2}}{r_{2}}$

                 = $\frac{9 \times 10^{9} \times 2.2 \times 10^{-6}}{0.4} + \frac{9 \times 10^{9} \times -3.20 \times 10^{-6}}{\sqrt{(0.4)^{2} + (0.4)^{2}}}$

                 = -4590 V

Therefore, we can conclude that the potential at point a, which is on the x axis at x = 40.0 cm is -4590 V.