what is the answer of part f? (7%) Problem 5: A positive charge of magnitude 01
ID: 1412741 • Letter: W
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what is the answer of part f?
(7%) Problem 5: A positive charge of magnitude 01 = 6.5 nC is located at the origin. A negative charge Q2 =-85 nC is located on the positive x-axis at x = 12.5 cm from the origin. The point P is located y = 12.5 cm above charge Q2. Qi Q2 ©theexpertta.com 17 % Part (a) Calculate the x-component of the electric field at port P due to charge 01 Write your answer in units of NC to charge Q1. Write your answer in units ofN/C. 4 17% Part (b) Calculate the y-component of the electric field at point P due to charge Q1 Write your answer in units of NC 17% Part c) Calculate the y component of the electric field at pont P due to the Charge Q Write your answer in units of NC 17% Part (d) Calculate the y component of the electric field at point P due to both charges. Write your answer in units ofNC q) 17% Part(e) Calculate the magnitude of the electric field at point due to both charges. Write your answer in units ofNC #17% Part (f) Calculate the angle in degrees of the electric field at point P relative to the positive x-axis. Grade SummaryExplanation / Answer
(f)
q1 = 6.5 nC
q2 = - 8.5 nC
x = 12.5 cm = 0.125 m
y = 12.5 cm = 0.125 m
r = sqrt(x^2 + y^2)
r = sqrt(0.125^2 + 0.125^2)
r = 0.177 m
X component of Electric Field at point p due to charge q1 =
E1x = k*q1/r^2 * 0.165/0.207
E1x = (8.9 * 6.5)/ 0.177^2 * 0.125/0.177 N/C
E1x = 1304 N/C
Y component of Electric Field at point p due to charge q1 =
E1y = k*q1/r^2 * 0.165/0.207
E1y = (8.9 * 6.5)/ 0.177^2 * 0.125/0.177 N/C
E1y = 1304 N/C
X component of Electric Field at point p due to charge q2 = 0
Y component of Electric Field at point p due to charge q2 =
E2y = k*q2/r^2
E2y = - (8.9 * 8.5)/ 0.125^2 N/C
E2y = -4842 N/C
Ex = 1304 N/C
EY = - 4842 + 1304 = -3538 N/C
= tan^-1(3538/1304)
= 69.8o Clockwise from +ve x axis !!
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