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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?

User Yadvendar
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5.4k points

2 Answers

6 votes

Final answer:

To make the total potential at the remaining empty corner of the square 0 V, a negative charge of -2 * 7.67μC / √2 should be placed at one of the empty corners.

Step-by-step explanation:

To determine the charge that should be fixed to one of the empty corners so that the total potential at the other empty corner is 0 V, we need to apply the principle of superposition for electric potentials. Since we have two identical +7.67 μC charges fixed at adjacent corners of a square, let's assume the side of the square is d. The electric potential due to a point charge is given by the formula V = k * Q / r, where k is Coulomb's constant, Q is the charge, and r is the distance from the charge to the point where potential is being calculated.

There are two contributions to the potential at the empty corner: the direct diagonal contribution from the charges V1 and V2, which are both k * 7.67μC / (√d * d). Since the potentials due to individual charges are additive, we have a total contribution from the two charges of 2V1. To make the total potential zero, we need to place a charge Q3 at one of the empty corners such that the potential it creates V3 cancels out the 2V1.

V3 must equal -2V1. Since V3 = k * Q3 / r and r in this case is equal to d, we can write k * Q3 / d = -2(k * 7.67μC / (√d * d)). Solving for Q3, we find that Q3 = -2 * 7.67μC / √2. The algebraic sign is negative, meaning that the charge must be negative to create an opposing electric potential that cancels out the positive potential.

User Raymond Seger
by
5.3k points
4 votes

Answer:

q₃ = - 13.0935 μC

Step-by-step 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

User Tnwei
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5.1k points