What is the strength of the electric field

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Homework Statement



734snr8.jpg


A. What is the strength of the electric field at the position indicated by the dot? (X=2.84 cm, s=5.37 cm, q=5.5 nC.)

B. What is the direction of the electric field at the position indicated by the dot? Specify the direction as an angle (in degrees) above or below horizontal.


Homework Equations





The Attempt at a Solution


A. Do I just do E = K*Q / R^2 = 13413.7
Or do I need to find components of each charge force?
 

Answers and Replies

  • #2
dynamicsolo
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The distance of both charges above or below the position of the dot is the same, so I'll bet you can answer part (b) right off the bat...

For part (a), you will want to look at the components of the electric field produced by each charge at the location of the dot. But what components can you ignore?
 
  • #3
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Ok so part be is 0 degrees...
Part A...I got...
A being the top charge and B the bottom...
Ax = 13413.7 cos45 = 9484.92
Bx = 13413.7 sin45 = 9484.92
Rx = 18969.84

Ay = 13413.7 cos45 = 9484.92
By = -13413.7 sin45 = -9484.92
Ry = 0

R = Sqrt 18969.84 = 137.7
 
  • #4
dynamicsolo
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Ok so part be is 0 degrees...
Good!

Part A...I got...
A being the top charge and B the bottom...
Ax = 13413.7 cos45 = 9484.92
Bx = 13413.7 sin45 = 9484.92
Rx = 18969.84
I agree with your magnitude for the electric fields at the position of the dot. But the angle those field vector make to the horizontal direction is not 45º. Also, if you are finding the horizontal components of the two fields, you wouldn't be using cosine for one and sine for the other. Draw in the actual right triangles to see what angles and trig functions you should be using.

Ay = 13413.7 cos45 = 9484.92
By = -13413.7 sin45 = -9484.92
Ry = 0
The vertical component for the total field will indeed be zero, but this only works because sin 45º = cos 45º. Again, though, 45º is not the correct angle and the vertical components of the fields will not involve both sine and cosine...

Incidentally, you don't actually need to find the value of the angle. You can consider instead what ratios of sides of the right triangles you need for the sines and cosines.
 
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  • #5
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Sorry big mistake on the sin/cos thing!!
As for the angle...
Tan = .0537/.0284 = 1.89 tan-1 = 62.13

Ax = 13413.7 cos62.13 = 6270.46
Bx = 13413.7 cos62.13 = 6270.46
Rx = 12540.92

Ay = 13413.7 sin62.13 = 6270.46
By = -13413.7 sin62.13 = -6270.46
Ry = 0

But I am not sure how to find the angle I guess...
 
  • #6
dynamicsolo
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Tan = .0537/.0284 = 1.89 tan-1 = 62.13

Ax = 13413.7 cos62.13 = 6270.46
Bx = 13413.7 cos62.13 = 6270.46
Rx = 12540.92

Ay = 13413.7 sin62.13 (OK) = 6270.46 (copying error?)
By = -13413.7 sin62.13 = -6270.46
Ry = 0
Looks fine! You can avoid having to know that angle if you have the distance you calculated from each charge to the dot; you would have found 6.07 cm. The x-components will then be

13414 · (2.84 cm/6.07 cm) = 6271 N/C for each charge, and the y-components will be

+/- 13414 · (5.37 cm/6.07 cm) = +/- 11867 N/C .

This way, you don't need to take time to find the angle the field vectors make to the horizontal (since you aren't asked for that).

But I am not sure how to find the angle I guess...
Which angle do you mean?

For your result for part (b), you find that there is a positive x-component and a zero y-component for the total field at the dot, so the total field has magnitude 12,541 N/C pointing directly to the right. However, as you correctly noted earlier, since you can see that the individual y-components will cancel out, the direction of the field will point to the right; you don't need to make a computation to answer that part.
 
  • #7
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So my answer would be....
sqrt 12540.91^2 = 12540.92?
 
  • #8
dynamicsolo
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So my answer would be....
sqrt 12540.91^2 = 12540.92?
Well, yes, you could say that for the magnitude. But since the total y-component is zero, the total x-component is going to be the magnitude for the total field; you don't need to run through a calculation.
 

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