Finding the magnitude and direction at a point in an electric field with 2 charges

  • Thread starter infected
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Homework Statement


"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?


Homework Equations


E= Q/(4PIε°*r^2)
ε°=8.85*10^-12

The Attempt at a Solution


Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C

Direction is 158 Degrees from the X(got this from the vectors)

This is my answer, my physics is really rusty and I have a gut feeling I've gone completely wrong. A checking and assistance would be appreciated.

Thanks.
 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi infected! welcome to pf! :smile:

(have a pi: π and try using the X2 and X2 buttons just above the Reply box :wink:
"A point charge of 3*10^-6 C is at the origin and another point charge of -4*10^06 C is at point 0.1m away along the x axis.
What is the magnitude and direction of the electric field at point 0.15m from the origin along the posting y-axis?

Electric Field from the first charge to the point on the Y axis:
E1 = (3*10^-6)/(4PIε°*0.15^2)

Electric Field from the Second charge to the point on the Y axis:
E2 = (-4*10^-6)/(4PIε°*0.18^2)

Magnitude Of Electric Field = E1+E2=88808N/C
(you mean the positive y-axis?)

your E1 and E2 magnitudes look ok :smile:

but i think you've added the magnitudes, instead of using vector addition (ie adding the coordinates separately, or using a vector triangle) :wink:
 

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