Question

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm. Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.(1)What is the electric potential at point a due to q1 and q2?

(2)What is the electric potential at point b?

(3)A point charge q3 = -2.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer 1

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

`V = (1)/(4\pi\epsilon_0)(q)/(r)`

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

`V_a = (1)/(4\pi\epsilon_0)(q_1)/(r_1) + (1)/(4\pi\epsilon_0)(q_2)/(r_2)`

The distance from the center of the square to one of the corners is
`\sqrt2 L/2 = 0.035m`

`V_a = (1)/(4\pi\epsilon_0)(2*10^(-6))/(0.035) + (1)/(4\pi\epsilon_0)(-2*10^(-6))/(0.035) = 0`

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2)
`V_b = (1)/(4\pi\epsilon_0)(q_1)/(r_1) + (1)/(4\pi\epsilon_0)(q_2)/(r_2)`

`r_1 = 0.05\sqrt2m\\r_2 = 0.05m`

`U = (1)/(4\pi\epsilon_0)(q_1q_3)/(r) = Vq_3`

`W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2*10^(-6))(-(2.9*10^(-6))/(\pi\epsilon_0))\\W = (5.8*10^(-12))/(\pi\epsilon_0)`