# Electric Field Lines and Equipotential Contours

## 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.

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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gneill
Mentor
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?

Equipotential surface + value of charge needed. Sounds like gauss's law to me.

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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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.

gneill
Mentor
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.

gneill
Mentor
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.

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.

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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i think just use V = kq/r. then solve for q.

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.

Whats the equation for e field between 2point charges? and whats the equation for potential between 2 point charges?

gneill
Mentor
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!