Two charges, Q1= 2.70 muC, and Q2= 7.00 muC are located at points (0,-4.00 cm )

Question

Two charges, Q1= 2.70 muC, and Q2= 7.00 muC are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone? What is the x-component of the total electric field at P? What is the y-component of the total electric field at P? What is the magnitude of the total electric field at P? Now let Q2 = Q1 = 2.70 muC. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P? Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P?

Explanation / Answer

1) the distance will be sqrt ( 42 +62) =7.211cm =0.071m

so , field due to Q1= KQ1/r2

=(9x109 * 2.7 x10-6) /(0.071)2

=4.82x106 N/C

2) Q1 creates a field that is up and to the right at angle of arctan(4/6) =33.69o

Q2 creates a field that is down and to the right at angle of arctan(-4/6) =-33.69o

Ex =KQ1/r2*cos(33.69) + KQ2/r2*cos(-33.69)

= (9x109 * 2.7 x10-6) /(0.071)2cos(33.69) + (9x109 * 7 x10-6) /(0.071)2*cos(-33.69)

=1.7282x107N/C

3) Ey=KQ1/r2*sin(33.69) + KQ2/r2*sin(-33.69)

= (9x109 * 2.7 x10-6) /(0.071)2sin(33.69) + (9x109 * 7 x10-6) /(0.071)2*sin(-33.69)

=-4.94x105N/C

4)c) mag = sqrt(Ex^2 + Ey^2) = sqrt((1.7282x107)2 + (-4.94x105)2) =1.729x10^7N/C

5) d) Now only the y component is non zero due to symmetry

So Ey = 2*KQ1/r2*sin(33.69) = 2* (9x109 * 2.7 x10-6) /(0.071)2sin(33.69) = 6.204x105N/C

6)
e) F = E*q = 6.204x105N/C *1.60x10-19C = 9.92x10-14N

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