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 determine the charge that should be fixed to one of the empty corners, such that the total potential at the remaining empty corner is 0 V, we can use the concept of electric potential and the formula for calculating the electric potential due to a point charge.

The formula for electric potential, φ, due to a point charge is given by:

φ = k • q / r

Where:
- φ is the electric potential
- k is the electrostatic constant (approximately equal to 9 × 10^9 Nm^2/C^2)
- q is the charge of the point charge
- r is the distance between the point charge and the point at which the potential is being calculated

In this problem, we have three charges fixed at the corners of a square: two charges with a magnitude of +2.83 μC (microcoulombs) each, and one empty corner.

To solve for the charge at the empty corner, we need to calculate the electric potential at that corner and set it equal to 0 V.

Let's designate the charges as q1, q2, and q3:

q1 = q2 = +2.83 μC (given charge)
q3 = charge at the empty corner (to be determined)

The distances between the charges and the potential point P (empty corner) are as follows:

Distance between q1 and P: a
Distance between q2 and P: a/√2 (since the charges are at adjacent corners of a square)
Distance between q3 and P: a (since they are at the same corner)

Now, we can apply the formula for electric potential to each charge:

φ1 = k • q1 / a
φ2 = k • q2 / (a/√2)
φ3 = k • q3 / a

Since the total potential at point P is 0 V, we can combine the potentials:

φ1 + φ2 + φ3 = 0

Substituting the values and simplifying:

k • q1 / a + k • q2 / (a/√2) + k • q3 / a = 0

The constants k and a are common to each term, so we can factor them out:

k (q1 / a + q2 / (a/√2) + q3 / a) = 0

Simplifying further:

q1 / a + q2 / (a/√2) + q3 / a = 0

Substituting the known values:

2.83 μC / a + 2.83 μC / (a/√2) + q3 / a = 0

To find q3, we isolate it on one side of the equation:

q3 / a = - (2.83 μC / a + 2.83 μC / (a/√2))

Multiplying both sides by a:

q3 = - (2.83 μC + 2.83 μC √2)

Now, we can calculate q3:

q3 = - (2.83 μC + 2.83 μC √2)
= - 2.83 μC (1 + 1/√2)

Therefore, 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, is -2.83 μC (1 + 1/√2).