Consider a square which is 1.0 m on a side. Charges are placed at the corners (m
ID: 2237858 • Letter: C
Question
Consider a square which is 1.0 m on a side. Charges are placed at the corners (marked A,B,C D) of the square as shown in the figure.(Figure 1) Q = 9 ?C and q = 4 ?C The distance of each corner to the center of the square is 0.707 m. Part A What is the magnitude of the electric field at the center of the square? Express your answer using three significant figures. Part B What is the direction of the electric field at the center of the square. (angle measured in degrees CCW from the positive x axis in the usual sense) Angle = degrees SubmitMy AnswersGive Up Part C What is the electric potential at the center of the square? Electric potential = Volts SubmitMy AnswersGive Up Part D If a charge is placed at the center of the square, what is the x-component of the electrostatic force on the charge? = N SubmitMy AnswersGive Up Part E If a charge is placed at the center of the square, what is the y-component of the electrostatic force on the charge? = N SubmitMy AnswersGive Up Part F If a charge is placed at the center of the square, how much work (positive or negative) would you have to do to remove it to infinity? Work = J SubmitMy AnswersGive Up Provide FeedbackContinue of 1Explanation / Answer
The electric fields due to the charges at (0, 0) and (1, 1) cancel each other out at the center of the square, so we need only consider the charges at (1, 0) and (0, 1). Since the field vectors due to those charges point in the same direction (because they are of opposite signs), we can just compute the field due to one of them and double it.
The formula for the magnitude of the field due to a charge q at distance r is
E = (1 / (4 pi epsilon-0))(q / r^2).
In this case, q is 3 * 10^(-6), and r is sqrt(2) / 2. The value of the constant epsilon-0 is 8.9 * 10^(-12) in units consistent with the others in the problem. So we compute:
E = 2 * (1 / 4 pi (8.9 * 10^(-12))) (3 * 10^(-6) / (1/2)) = 1.07 * 10^5 N/C