Consider a square which is 1.0 m on a side. Charges are placed at the corners (m

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 = 10 C and q = 2 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)

Part C

What is the electric potential at the center of the square?

Part D

If a 1C charge is placed at the center of the square, what is the x-component of the electrostatic force on the charge?

Part E

If a 1C charge is placed at the center of the square, what is the y-component of the electrostatic force on the charge?

Part F

If a 1C 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?

(THANK YOU FOR TAKING THE TIME TO ANSWER THIS!)

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Explanation / Answer

a)

r = distance of each charge from centre = 0.707 m

At the center electric field due to charge Q placed at A and C cancels out being equal and opposite in direction

the electric field by charges at B and D is in same direction and equal in magnitude

hence net electric field at centre is given as

E = k q/r2 + k q/r2 = 2 (9 x 109) (2 x 10-6)/(0.707)2 = 7.2 x 104 N/C

b)

direction : electric field is towards the charge located at D

angle = 90 + 45 = 135 CCW

c)

charge at B and D are equal in magnitude and opposite in nature and are at same distance from center. hence they produce equal and opposite electric potential at center which add up to 0

Total electric potential at center

V = k Q/r + kQ/r = 2 kQ/r = 2 (9 x 109) (10 x 10-6)/(0.707) = 2.55 x 105 volts

d)

electric field is towards D . if we place a negative charge at the center it will experience electric force in opposite direction of electric field which is towards B .

F = magnitude of electric force = q' E = (1 x 10-6) (7.2 x 104) = 0.072 N

X-component is given as

Fx = F Cos45 = 0.072 Cos45 = 0.051 N

E)

Y-component of force is given as

Fy = - F Sin45 = - 0.072 Cos45 = - 0.051 N

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