A positive charge of magnitude Q1 = 7.5 nC is located at the origin. A negative

Question

A positive charge of magnitude Q1 = 7.5 nC is located at the origin. A negative charge Q2 = -4.5 nC is located on the positive x-axis at x = 7.5 cm from the origin. The point P is located y = 9.5 cm above charge Q2.

1) Calculate the x-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

2) Calculate the y-component of the electric field at point P due to charge Q1. Write your answer in units of N/C.

3) Calculate the y-component of the electric field at point P due to the Charge Q2. Write your answer in units of N/C.

4) Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C.

5) Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C.

6) Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.

Explanation / Answer

Solution:

a) Ex1 = kQ1 / R^2 ( cos theta)

Ex1 = k Q1 ( x / sqrt(x^2 + y^2) / (x^2 + y^2)

Ex1 = kQ1x / (x^2 + y^2)^1.5

Ex1= (9 x 10^9 x 7.5 x 10^-9 x 0.075 ) / (0.075^2 + 0.095^2)^1.5

Ex1 = 2855.02 N/C

b) Ey1 = kQ1 / R^2 ( sint heta)

Ey1 = k Q1 ( y / sqrt(x^2 + y^2) / (x^2 + y^2)

Ey1 = kQ1y / (x^2 + y^2)^1.5

Ey1= (9 x 10^9 x 7.5 x 10^-9 x 0.095 ) / (0.075^2 + 0.095^2)^1.5

Ey1 = 3616.75 N/C

c) Ey2 = - kQ2 / y^2

Ey2 = - (9 x 10^9 x 4.5 x 10^-9) / (0.095^2) = - 4487.53 N/C

d) EY = Ey1 + Ey2 = - 870.78 N/C

e) magnitude = sqrt(Ey^2 + Ex^2) = 2984.86 N/C

f) theta = tan^-1(870.78/ 2855.02) = 16.96 deg below x axis

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