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

6/12=1/2

step-by-step explanation:

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