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.

Answer 1

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

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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

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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

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angle = tan^-1(Ey/E1x) = 81.8 <<<--------------answer