Identical +2.83 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?
φ =k•q/r
φ1+ φ2 + φ3 =0
k•q1/a + k•q2/a√2 + k•q3/a =0
q3= q1 + q2/√2 =2.83(1+1/√2) =
=
To find the charge (magnitude and algebraic sign) that should be fixed to one of the empty corners, so that the total potential at the remaining empty corner is 0 V, you can follow these steps:
1. Start with the formula for electric potential, φ = k•q/r, where φ is the potential, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
2. Consider the potential at each corner of the square. Let's label them as corners 1, 2, 3, and 4.
3. The potential at corner 4 should be 0 V, so the sum of the potentials at corners 1, 2, and 3 must add up to 0 V. We can write this as φ1 + φ2 + φ3 = 0.
4. From the formula for electric potential, we know that the potential at a corner is proportional to the charge and inversely proportional to the distance. Therefore, we can rewrite the equation as k•q1/a + k•q2/a√2 + k•q3/a = 0, where a is the distance between two adjacent corners.
5. Now, substitute the charge values into the equation. Let's assume q1 and q2 are the identical +2.83 μC charges fixed to the adjacent corners. Therefore, q1 = q2 = 2.83 μC.
6. Substitute the values into the equation: k•(2.83 μC)/a + k•(2.83 μC)/(a√2) + k•q3/a = 0.
7. Simplify the equation: 2.83k/a + 2.83k/(a√2) + k•q3/a = 0.
8. Combine the terms with the same denominator: 2.83k(1/a + 1/(a√2)) + k•q3/a = 0.
9. We want to find q3, the charge at the remaining empty corner. From the equation, we can see that q3 = -2.83(1 + 1/√2) μC.
Therefore, the charge (magnitude and algebraic sign) that should be fixed to one of the empty corners is -2.83(1 + 1/√2) μC.