(a) Three point charges, A = 2.2 ?C, B = 7.5 ?C, and C = -3.9 ?C, are located at
ID: 2167526 • Letter: #
Question
(a) Three point charges, A = 2.2 ?C, B = 7.5 ?C, and C = -3.9 ?C, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 2.2 ?C charge.
magnitude N/C
direction ? (counterclockwise from the +x-axis)
(b) How would the electric field at that point be affected if the charge there were doubled? The magnitude of the field would be halved. The field would be unchanged. The magnitude of the field would double. The magnitude of the field would quadruple.
(c) Would the magnitude of the electric force be affected? no yes
Explanation / Answer
Three point charges, A = 2.2 µC, B = 7.5 µC, and C = -3.9 µC, are located at the corners of an equilateral triangle as in the figure above. Find the magnitude and direction of the electric field at the position of the 2.2 µC charge.
figure shows a triangle with point A having a 60° angle and a positive charge. point B with a positive charge and 0.5m away from point C which has a negative charge. Point A is the bottom left of triangle, point B is at the top, and point C is at bottom right
An equilateral triangle has all 3 sides equal and all 3 angles equal.
Since side BC = 0.5 m,
Side AC = 0.5 m
Side AB = 0.5 m
If we say that Point A is the origin, Point A = (0 ,0)
Point C is 0.5 m to the right, so Point C = (0.5 , 0)
Line AB is 0.5 m long and angle CAB = 60º
Force = (9 * 10^9 * q1 * q2) ÷ r^2
The distance from Point B to Point A = 0.5 m
Force = (9 * 10^9 * 7.5 * 10^-6 * 2.2 * 10^-6) ÷ 0.5^2
Force = 0.594 N
The distance from Point C to Point A = 0.5 m
Force = (9 * 10^9 * -3.9 * 10^-6 * 2.2 * 10^-6) ÷ 0.5^2
Force = 0.3089 N in the negative x direction
The force that the 7.5 µC charge at Point B exerts on the 2.2 µC at Point A = 0.594 N
The vertical component of the force that the 7.5 µC charge at Point B exerts on the 2.2 µC at Point A = 0.594 * sin 60º in the negative y direction.
The horizontal component of the force that the 7.5 µC charge at Point B exerts on the 2.2 µC at Point A = 0.594 * cos 60º negative x direction.
The total horizontal component of the force on Point A =
-0.3089 N + -0.594 * cos 60º = -0.6059 N
The total vertical component of the force on Point A =
-0.594 * sin 60º = -0.5144 N
The total force at the position of the 2.2 µC charge =
(-0.6059^2 + -0.5144^2)^0.5 = 0.7948 N
The tangent of the angle = -0.5144 ÷-0.6059
Angle = tan^-1 (-0.5144 ÷-0.6059) = 40.33º
This angle is measured counter clockwise from the –x axis
So the angle counter clockwise from the +x axis = 40.33º + 180º= 220.33º
The magnitude and direction of the electric field at the position of the 2.2 µC charge = 0.7948 N at 220.33º counter clockwise from the +x axis
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