PHYS HELP Problem 8: In the figure, the point charges are located he corners of
ID: 1267908 • Letter: P
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PHYS HELP
Problem 8: In the figure, the point charges are located he corners of an equilateral triangle 25 cm on a side. qa Randomized Variables 25 cm Part (a Find the magnitude of the electric field in NC at the location of qa, given that qb 11.7 Hc and qc --4.1 uc Numeric A numeric value is expected and not an expression Part (b Find the direction of the electric field at qa in degrees above the ne with origin at ga. Numeric A numeric value is expected and not an expression Part (e) What is the magnitude of the force in N on qa, given that qa 1.2 nc? Numeric A numeric value is expected and not an expression Part (d) What is the direction of the force on qa in degrees above the ne ive x-axis with origin at ga? Numeric A numeric value is expected and not an expression.Explanation / Answer
Part A)
Start with the formula for Electric Field...E = kq/r2
From qb...
E = (9 X 109)(11.7 X 10-6)/(.25)2
E = 1.68 X 106 N/C up and to the left
The x component is (1.68 X 106)(cos 60) = 8.42 X 105 N/C left
The y component is (1.68 X 106)(sin 60) = 1.46 X 106 N/C up
From qc...
E = (9 X 109)(4.1 X 10-6)/(.25)2
E = 5.90 X 105 N/C down and to the left
The x component is (5.90 X 105)(cos 60) = 2.95 X 105 N/C left
The y component is (5.90 X 105)(sin 60) = 5.11 X 105 N/C down
The net x = 8.42 X 105 + 2.95 X 105 = 1.14 X 106 N/C left
The net y = 1.46 X 106 - 5.11 X 105 = 9.49 X 105 N/C up
The net field is from the Pythagorean Theorem
net2 = (9.49 X 105)2 + (1.14 X 106)2
net = 1.48 X 106 N/C
Part B)
The direction is from the tangent formula
tan(angle) = 9.49 X 105/1.14 X 106
angle = 39.8o above the negative x axis
Part C)
F = qE
F = (1.2 X 10-9)(1.48 x 106)
F = 1.78 X 10-3 N
Part D)
The angle does not change from part B. It is still 39.8o above the negative x axis
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