Find the electric field intensity at point D

  • Thread starter aeromat
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  • #1
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1. The problem statement, all variables and given/knownThe diagram shows three small charges at three corners of a rectangle; charge A is -12µC, charge B is -15µC and charge C is +8.1µC. Calculate the magnitude and direction of the electric field intensity at the fourth corner, D.


The Attempt at a Solution


I actually did this entire question with components and got the right answer. However, I find that it is time-consuming, and I was wondering if any of you PF experts know a faster method of doing this.

Components Method:
Split every diagonal vector to x and y vectors and add up all x and y vectors from the three electric field intensity vectors in the image to get the Net[X] and Net[Y] vector. Then use Phythagorean Theorem to find the electric field intensity vector at D, and then use arctan(- / - ) to calculate the angle.

Is there a faster way of accomplishing this?
Thanks,
aeromat.
 

Answers and Replies

  • #2
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Well you really don't have to break it up into components, you can just use vector addition. Its really the same thing but it might save a bit of time. For example if you found the three electric field vectors due to the charges to be

[tex] \vec E_1 = e_{1x} \hat i + e_{1y} \hat j [/tex]

[tex] \vec E_2 = e_{2x} \hat i + e_{2y} \hat j [/tex]

[tex] \vec E_3 = e_{3x} \hat i + e_{3y} \hat j [/tex]

then you can add these to give the resultant E vector

[tex] E_r = (e_{1x} + e_{2x} + e_{3x} ) \hat i + (e_{1y} + e_{2y} + e_{3y} ) \hat j [/tex]

And then use the same procedure to get the intensity and angle.
It may save a bit of time, and it is good to get accustomed to using vectors.
 

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