Three point charges with values = 4.00 , = 1.00 , and = 7.00 are placed on three consecutive corners of a square whose side measures = 5.00 . Point A lies at the fourth corner of the square, diagonally across from . Point B lies at the center of the square. What are the values of the electric potential at point A and at point B? How much energy (supplied as work) is required to move a test charge = 2.00 from point A to point B? Assume the zero of electric potential is chosen to be at infinity.
To find the electric potential at point A, we can use the formula for electric potential due to a point charge:
V = k*q/r,
where V is the electric potential, k is the electrostatic constant (9 x 10^9 N*m^2/C^2), q is the charge, and r is the distance from the charge.
First, let's calculate the electric potential at point A due to each of the three charges. The values of the charges are given as 4 C, 1 C, and 7 C.
For the charge q1 = 4 C, the distance (r1) between point A and the charge is the length of the square's side, which is 5 m.
V1 = (9 x 10^9 N*m^2/C^2) * (4 C) / (5 m) = 7.2 x 10^9 V
Similarly, for the charge q2 = 1 C, the distance (r2) between point A and the charge is also 5 m.
V2 = (9 x 10^9 N*m^2/C^2) * (1 C) / (5 m) = 1.8 x 10^9 V
For the charge q3 = 7 C, the distance (r3) between point A and the charge can be calculated using the Pythagorean theorem, as it is diagonally across from point A.
r3 = sqrt((5 m)^2 + (5 m)^2) = 7.07 m
V3 = (9 x 10^9 N*m^2/C^2) * (7 C) / (7.07 m) = 9.9 x 10^9 V
To find the total electric potential at point A, we add up the contributions from each charge:
V_total_A = V1 + V2 + V3 = 7.2 x 10^9 V + 1.8 x 10^9 V + 9.9 x 10^9 V = 18.9 x 10^9 V
Therefore, the electric potential at point A is 18.9 x 10^9 V.
Now, let's calculate the electric potential at point B, which is the center of the square.
Since point B is equidistant from all the charges, the electric potentials due to each charge will be the same as the average of V1, V2, and V3:
V_avg = (V1 + V2 + V3) / 3 = (7.2 x 10^9 V + 1.8 x 10^9 V + 9.9 x 10^9 V) / 3 = 6.3 x 10^9 V
Therefore, the electric potential at point B is 6.3 x 10^9 V.
To calculate the work required to move a test charge from point A to point B, we can use the formula:
W = q * (ΔV),
where W is the work done, q is the charge, and ΔV is the change in electric potential.
In this case, the charge, q, is given as 2 C.
The change in electric potential, ΔV, is the electric potential at point B minus the electric potential at point A:
ΔV = V_B - V_A = 6.3 x 10^9 V - 18.9 x 10^9 V
ΔV = -12.6 x 10^9 V
Now, we can calculate the work:
W = q * ΔV = 2 C * (-12.6 x 10^9 V) = -25.2 x 10^9 J
Therefore, the work required to move the test charge from point A to point B is -25.2 x 10^9 J (negative because work is done against the electric field).