Identical + 7.67 μC charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total potential at the remaining empty corner is 0 V?

q₃ = - 13.0935 μC

Explanation:

Given

q₁ = q₂ = + 7.67 μC

We use the equation

V = Kq/r

We can apply it as follows

V₁ = K*q₁/r₁ = K*q₁ / (√2*L)

V₂ = K*q₂/r₂ = K*q₂/L

V₃ = K*q₃/r₃ = K*q₃/L

Then

V₁ + V₂ + V₃ = 0

⇒ (K*q₁ / (√2*L)) + (K*q₂/L) + (K*q₃/L) = 0

⇒ (K/L) * ((q₁/√2) + q₂ + q₃) = 0

⇒ (q₁/√2) + q₂ + q₃ = 0

Since q₁ = q₂

⇒ (q₁) ((1/√2) + 1) + q₃ = 0

⇒ q₃ = - (q₁) ((1/√2) + 1) = + 7.67 μC * (1.7071)

⇒ q₃ = - 13.0935 μC