What are the strength and direction of the electric field at the position indica
ID: 2125167 • Letter: W
Question
What are the strength and direction of the electric field at the position indicated by the dot in the diagram below, in which d = 4.9 cm, q = 8 nC, the positive x-axis points to the right, and the positive y-axis points up? Give your answer in each of the following forms.
What are the strength and direction of the electric field at the position indicated by the dot in the diagram below, in which d = 4.9 cm, q = 8 nC, the positive x-axis points to the right, and the positive y-axis points up? Give your answer in each of the following forms.Explanation / Answer
Part A)
From the q directly to the right
E = (9 X 10^9)(8 X 10^-9)/(.049)^2 = 29988 N/C to the left
From the 5nC charge directly below
E = (9 X 10^9)(5 X 10^-9)/(.02)^2
E = 112500 N/C straight down
From the charge in the corner
The distance... r^2 = (.049)^2 + (.02)^2 = .0028
E = (9 X 10^9)(8 X 10^-9)/(.0028)
E = 25705 N/C
This one needs components...
the angle is from the tangent function
tan (angle) = 2/4.9
angle = 22.2 degrees
The x component is 25705(cos 22.2) = 23800 N/C to the left
The y component is 25707(sin 22.2) = 9713.2 N/C upwards
The net x = 29988 + 23800 = 53788 N/C to the left
The net y = 102787 N/C downward
In component form...
-53788i - 102787j
Part B)
For the net...
net^2 = (53788^2) + (102787)^2
net = 116010 N/C
The direction... tan(angle) = 102787/53788
angle = 62.4 degrees
From the x axis that is 180 + 62.4 = 242.4 degrees
In summary...
116010 N/C at 242.4 degrees
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