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Electric Field Lines and Equipotential Contours

  • Thread starter jaguar7
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Homework Statement



The adjacent figure shows the E-field lines (with arrows) and equipotential contours for a certain charge configuration. Determine for each of the following statements whether it is true or false.

2117d6a269deae05065bfd232588bab843e47e025f6cb2214dd17e793819e4413e0d22cc2c98ef4f0c8636a20fdbbc1b.gif


Assuming that these contours are those of a point charge at the origin, calculate the value of the charge.

Homework Equations



Instructor says "Yes - read off the "r" values right from the diagram.

I don't think you'll need to integrate anything to solve this one. (Unless you feel like repeating the integral we already solved for a point charge.)

Also - looks like the diagram is assuming V is zero at r = infinity."


The Attempt at a Solution



okay, the first contour line of -9000V is at 1.00 meters and the third of -7000V is at 1.24 meters.

I use dV = kQ/rf - kQ/ri = kQ ( 1/rf - 1/ri)
Q = 2000 / (8.99E9 * (1/1 - 1/1.24))

then I switch it to negative since the E-field lines are pointing toward the charge

<the question is what am i doing wrong... grr...>
 
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Answers and Replies

  • #2
gneill
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Was there a question behind this? If not, I have one. If it's a point charge, the potential at a single distance should be able to give you the charge. It looks as though the -9000V potential at 1.0m is the most easily obtained with any accuracy from the diagram, so why not go with that?
 
  • #3
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Equipotential surface + value of charge needed. Sounds like gauss's law to me.
 
  • #4
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Equipotential surface + value of charge needed. Sounds like gauss's law to me.
Can we use Gauss's law without the E field? I don't think so. We have electric potential. V = E * r --- I don't know...?

I don't think if V = Er it would help us???
 
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  • #5
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E is -del V = -dV/dR.

E = (9000-6000V)/0.4 = 7500 N/C

Checking back to the graph, it seems true if the 7000V line is at 1.27m and 8000V line at 1.133m. (which seems quite reasonable comparing to the graph) unless u already know exactly where the lines are placed, this approach seems fine.
 
  • #6
gneill
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Going via Gauss' Law would be the long way round to the same thing. First you would find the form of the electric field produced by a point charge, then integrate it along a radial path to find the potential at distance r from the charge. Which we already have an expression for:

V = k*q/r^2

It's plug and play from this point. Pick a distance and potential (V). Solve for q.
 
  • #7
gneill
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E is -del V = -dV/dR.

E = (9000-6000V)/0.4 = 7500 N/C

Checking back to the graph, it seems true if the 7000V line is at 1.27m and 8000V line at 1.133m. (which seems quite reasonable comparing to the graph) unless u already know exactly where the lines are placed, this approach seems fine.
At exactly what radius will you assign this value of the E field? The field is dropping as 1/r^2 (you get a hint of this in the spacing of the equipotential lines in the figure). You've taken an average over a linear distance for a field that varies as 1/r^2, and the charge that you calculate is going to be pretty sensitive to the distance chosen.
 
  • #8
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Ah, sorry you're right, im assuming a constant E in this case. Didn't think to use the equation for point charge cause the top part of the post said it was a charge configuration. Just read the line below the picture and it's actually just a point charge.
 
  • #9
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Going via Gauss' Law would be the long way round to the same thing. First you would find the form of the electric field produced by a point charge, then integrate it along a radial path to find the potential at distance r from the charge. Which we already have an expression for:

V = k*q/r^2

It's plug and play from this point. Pick a distance and potential (V). Solve for q.
It's not V = kq / r^2.

It's E = kq / r^2... for Electric field. But that doesn't have voltage anywhere...

We don't know the value of the electric field anywhere I don't think...
 
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  • #10
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i think just use V = kq/r. then solve for q.
 
  • #11
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i think just use V = kq/r. then solve for q.
That's not a correct equation. It's not V = kq/r

I'm pretty sure that won't work - that is the equation for electric field, not electric potential.
 
  • #12
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Whats the equation for e field between 2point charges? and whats the equation for potential between 2 point charges?
 
  • #13
gneill
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That's not a correct equation. It's not V = kq/r

I'm pretty sure that won't work - that is the equation for electric field, not electric potential.
No, it's correct. Electric potential in volts is given by kq/r. Check the units!
 

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