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Electric Field Q

  • Thread starter t_n_p
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  • #1
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Homework Statement

Two charges -1nC and +1nC are positioned at (-2,0) and (2,0) respectively, where distances are in cm.

a) Sketch 5-6 field lines E in the region bounded by x=-2, x=2, y=0 and y=8

b) Caclculate the horizontal and verticle components of the E field at (-2,6), Draw a diagram showing the charges, the fields at P due to both charges and the resultant field. Find magnitude E and the angle phi (measured from the +x-axis)

The attempt at a solution

a)http://img241.imageshack.us/img241/1310/untitledxg6.jpg [Broken]

b) I'm not sure how to calculate the hotizontal and ver components of the E at (-2,6)!
 
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Answers and Replies

  • #2
Doc Al
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b) I'm not sure how to calculate the hotizontal and ver components of the E at (-2,6)!
What's the electric field at that point due to each point charge? Add the two fields--they are vectors--to get the net field at that point. (Hint: Find and add the components.)
 
  • #3
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Do I use coulombs law, |F| = (k)(|q1|)(|q2|)/(r^2)

and then E = F/Qo ?

I found the second formula in my notes, but not sure how Qo comes into it?
 
  • #4
Doc Al
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44,877
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Do I use coulombs law, |F| = (k)(|q1|)(|q2|)/(r^2)

and then E = F/Qo ?
Yes.

I found the second formula in my notes, but not sure how Qo comes into it?
Let q1 be your point charge (for which you are finding the field) and q2 be a test charge at point P. q2 = Qo.

When all is said and done, the magnitude of the field from a point charge Q is:
|F| = k|Q|/(r^2)
 
  • #5
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P doesn't appear to have a charge, so would I use
|F| = (k)(|q1|)(|q2|)/(r²)
or
|F| = k|Q|/(r²)?
 
  • #6
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I think I pretty much managed to do it..
Basically I used formula E = kq/r2 to find the E on P by point A(2,0) and the force on P by point B(-2,0).

Then I found the x and y components of the E on P by A.
Then calculated resultant in x direction = 1025.95V/m (this is equal to the E on P by A as E on P by B does not contribute to x direction.)

and

resultant in y direction = 2497.22(E on P by B in y direction) - 1521.03 (E on P by A in y direction)

Then I found magnitude (1416.17V/m) by pythag and angle using trig (44deg)
Heres a diagram, I know its messy

http://img402.imageshack.us/img402/83/img0351afn1.jpg [Broken]
 
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  • #7
lightgrav
Homework Helper
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okay, except that phi is supposed to be from the +x -axis.

Watch your language, though ...
the E-field is caused by Charges, not points,
and the E doesn't "act on" a point (only acts on a charge).
 
  • #8
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okay, except that phi is supposed to be from the +x -axis.

Watch your language, though ...
the E-field is caused by Charges, not points,
and the E doesn't "act on" a point (only acts on a charge).
so phi = 180+44 = 224deg?
 
  • #9
learningphysics
Homework Helper
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I'm getting different numbers. For the x-component due to the +1 n C... I get

[tex]\frac{kq}{r^2} = \frac{(9*10^9)(1*10^{-9})}{52}[/tex]

So that comes out to 9/52 = 0.173 V/m

then take the x - component... [tex]\frac{4}{\sqrt{52}}*0.173[/tex] which comes out to 0.096 V/m.

And for the y - component [tex]\frac{6}{\sqrt{52}}*0.173[/tex] which comes out to 0.144 V/m

I'm just giving the magnitudes, I'm ignoring the signs here...
 
  • #10
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I was partially wrong, but I think you are too. (?)

Units are in cm, so root(52)=0.07cm
0.07^2 = 0.052

But im confused. Do you convert to m after squaring the square root of (52) or before?
 
  • #11
learningphysics
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I was partially wrong, but I think you are too. (?)

Units are in cm, so root(52)=0.07cm
0.07^2 = 0.052

But im confused. Do you convert to m after squaring the square root of (52) or before?
Oops. Sorry about that.

I made a mistake earlier. You should convert before squaring. You meant to write 0.07m right?
 
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  • #12
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yes lol.
so radius should actually be..

root (52) = 7.211cm
7.211^2 = 52cm
52cm = 0.52m
=17.29V/m

But the other calculation is right for Epb
((8.99*10^9)*(1*10^-9))/(0.06^2) = 2497.22V/m.

One is so big, one is so small, could this be right?
 
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  • #13
lightgrav
Homework Helper
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distances are in centi-meters ... not meters, so
9e9 [Vm/C]*1e-9[C]/.0052[m^2] = 1731 V/m ,
. . . yielding an x-component of 960 V/m ... hmm. roundoff error?
 
  • #14
595
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How did you manage to get 0.0052?
(sqrt(52))²=52cm=0.052m?
 
  • #15
learningphysics
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yes lol.
so radius should actually be..

root (52) = 7.211cm
7.211^2 = 52cm
52cm = 0.52m
=17.29V/m

But the other calculation is right for Epb
((8.99*10^9)*(1*10^-9))/(0.06^2) = 2497.22V/m.

One is so big, one is so small, could this be right?
Sorry... I made a mistake in my post earlier that I edited afterwards. You should definitely convert to m before squaring.
 
  • #16
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Sweet, thanks! Finally got that all sorted..

The x component then becomes 966.76 V/m
and y component then becomes -1433.28V/m

Resultant thus becomes
x direction = 966.76 V/m
y direction = 2497.22-1433.28 = 1063.94

and magnitude becomes 1437.57V/m.

My phi now changes to 180+48=228deg measured from positive x axis
 
  • #17
learningphysics
Homework Helper
4,099
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Sweet, thanks! Finally got that all sorted..

The x component then becomes 966.76 V/m
and y component then becomes -1433.28V/m

Resultant thus becomes
x direction = 966.76 V/m
y direction = 2497.22-1433.28 = 1063.94

and magnitude becomes 1437.57V/m.

My phi now changes to 180+48=228deg measured from positive x axis
Cool! remember to convert your units before you do any squaring or anything... sorry about my earlier blunder.
 
  • #18
595
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Cool! remember to convert your units before you do any squaring or anything... sorry about my earlier blunder.

Not to worry! Thanks alot for helping in this thread and my other one! *bows to physics god*
 

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