# Determine the point at which the electric field is equal to zero?

Two point charges, -2.5 micro coulombs and 6 micro coulombs, are separated by a distance of 1m (with the -2.5 charge on the left and 6 on the right).

What is the point where the electric field is zero?

This seems exceptionally easy, but I can't figure it out. I can calculate the answer if both charges are negative/positive easily, but the fact that they're different is confusing me.

Thanks!

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Hint: What would be the direction of the electric field, say $100m$ away? –  Qmechanic May 14 '11 at 11:18
As a more general problem solving technique, have you drawn a diagram? –  Nic May 17 '11 at 9:51

Because of symmetry, the point at which the forces cancel must be on the same line as the two charges -- it's effectively a 1-dimensional problem.

If a small positive test charge were between the two charges, it would be "pushed" or "pulled" in the same direction by both charges, so it cannot be there.

If a small positive test charge were to the right of the two charges, it would always be "pushed" harder to the right by the bigger charge than it would be "pulled" to the left by the smaller charge, so it cannot be there.

A small positive charge that is placed to the left will be pulled to the right by the nearby negative charge and pushed to the left by more distant, but larger positive charge, so the balance point must be somewhere here.

So, taking the force that pushes or pulls a positive test charge $q$ at a distance $x$ to the right of the negative charge to be positive, there is a force proportional to $2.5q/x^2$ due to the nearby negative charge and a force proportional to $-6q/(1+x)^2$ due to the more distant positive charge. Adding the two together and asking the sum to be zero gives you an equation that you can convert into a quadratic equation. You only want the positive solution, right? Why?

What's the key to getting this kind of thing right? Decide what direction you're going to call positive and stick to it ruthlessly. Introducing a test charge and choosing it to be positive helps here by making the question less abstract. Draw a rough diagram to help you keep which is which straight in your head. I had to be careful to make sure that choosing a positive charge instead of a negative charge doesn't make a difference (check why not, why it helped me draw a diagram, and what the diagram would look like if I were to use a negative charge).

This homework question is too easy for Physics SE. Ask a different kind of Question next time. It would have been better if you had set out carefully and in more detail in your Question why the approach you took didn't work out. There's a good chance that if you had set it out carefully you would have seen why you had a problem with the calculation and the answers you were getting.

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Following Peter Morgan's eqn, I worked out the answer to be ABOUT (I did not do this on a spreadsheet) (13/7) meters on a straight line away from the smaller charge and opposite the larger charge. That is, the smaller charge is between the zero point and on a straight line from the zero point and the larger charge.

One way to visualize what is happening: Think of the earth have a gravity field of -2.5/(x^2), where x is your height above earth, measured is AU's (distance of Earth to Sun). So, the earth is going to repel you, and you are floating out in space. Now assume that the Sun is pulling you toward it with a force of 6/(z^2). If you fall off the Earth in any other direction other than straight away from the Sun, you will be pulled into the Sun. But if you are lucky enough to fall off the Earth directly away from the Sun, you will reach a point where the Earth's repulsion of you is exactly countered by the Sun's pull. It would be a Lagrangian point of sorts (the unstable sort, actually). That point would be a little less than (13/7) AUs away from the Earth, and a little less than (20/7) AUs away from the Sun.

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