A positive charge of magnitude Q 1 = 9 nC is located at the origin. A negative c
ID: 1397642 • Letter: A
Question
A positive charge of magnitude Q1 = 9 nC is located at the origin. A negative charge Q2 = -9 nC is located on the positive x-axis at x = 18.5 cm from the origin. The point P is located y = 8.5 cm above charge Q2.
Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.
Calculate the y-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.
Calculate the y-component of the electric field at point P due to the Charge Q2. Write your answer in units of N/C.
Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C.
Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C.
Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.
Explanation / Answer
q1 = 9 nc r1 = sqrt(8.5^2+18.5^2) = 20.4 cm = 0.204 m
q2 = -9nC r2 = 8.5 cm = 0.085 m
tan theta = 8.5/18.5
theta = 26.7 degrees
x-component of the electric field at point P due to charge Q1.
E1x = k*q1*cos26.7/r1^2 = (9*10^9*9*10^-9*cos26.7)/(0.204^2) = 1738.8 N/C <<<--------------answer
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y-component of the electric field at point P due to charge Q1
E1y = k*q1*sin26.7/r1^2 = (9*10^9*9*10^-9*sin26.7)/(0.204^2) =
874.5 N/C <<<--------------answer
----------------------
y-component of the electric field at point P due to the Charge Q2.
E2y = -k*q2/r2^2 = -(9*10^9*9*10^-9)/(0.085^2) = -11211.07 N/C <<<--------------answer
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y-component of the electric field at point P due to both charges
Ey = E1y + E2y = 12085.57 N/C <<<--------------answer
---------------------
magnitude of the electric field at point P due to both charges
E = sqrt(Ey^2+E1x^2)
E = sqrt(1738.8^2+12085.57^2)
E = 12210.01 N/C <-------answer
----------
angle = tan^-1(Ey/E1x) = 81.8 <<<--------------answer
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