Identical +3.92 ìC 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?
To find the charge that should be fixed to one of the empty corners of the square such that the total potential at the remaining empty corner is 0 volts, we can use the principle of superposition.
The potential due to a single point charge can be calculated using the formula:
V = k * q / r
where V is the potential, k is the Coulomb's constant (9 * 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
In this case, we have identical charges (+3.92 μC) fixed to two adjacent corners of the square. The distance between the charges is equal to the length of one side of the square.
Let's assume that the charge to be fixed to the empty corner is q2. The potential at the remaining empty corner due to these charges can be expressed as:
V_total = k * q1 / r1 + k * q2 / r2
Since we want the total potential at the remaining empty corner to be zero (V_total = 0), we can write the equation as:
0 = k * q1 / r1 + k * q2 / r2
Given that the charges q1 = q2 = 3.92 μC, and the distances r1 = r2 = side length of the square, we can substitute these values into the equation:
0 = k * (3.92 * 10^-6 C) / r1 + k * (q2) / r2
Simplifying the equation, we get:
0 = (3.92 * 10^9 Nm^2/C^2) * (3.92 * 10^-6 C) / r1 + (3.92 * 10^9 Nm^2/C^2) * (q2) / r2
0 = (15.3664 / r1) + (q2 / r2)
To make the total potential zero, the value of q2 / r2 should be equal to -15.3664 / r1.
Since the distance between the charges on the square is equal to the side length of the square, we can simplify the equation further:
0 = (15.3664 / r) + (q2 / r)
To isolate q2, we can move the (15.3664 / r) term to the other side of the equation, which gives:
q2 / r = -(15.3664 / r)
Multiplying both sides of the equation by r gives:
q2 = -15.3664
So, the magnitude of the charge that should be fixed to the empty corner is 15.3664 μC, and the algebraic sign is negative (-).
φ =k•q/r
φ1+ φ2 + φ3 =0
k•q1/a + k•q2/a√2 + k•q3/a =0
q3= q1 + q2/√2 =3.92(1+1/√2) =
= 6.69 μC