# The Electric field between two charges

1. Jan 28, 2014

### K.QMUL

1. The problem statement, all variables and given/known data

Use a Cartesian system whose origin coincides with the midpoint between the two charges and write
the expression for the magnitude of electric field $\vec{E}$
on the line passing through the two charges.

Then Show that when the two charges have the same sign, there is a point X0 along the line connecting the two charges where $\vec{E}$ vanishes

2. Relevant equations

E=$\frac{KQ}{r^2}$

3. The attempt at a solution

Please look at the attachment. I am unsure if I have done it right.

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2. Jan 28, 2014

### haruspex

Signs are important here. It is wrong to take the modulus of the charges before combining them. When the charges have the same sign the field will be different from when they are opposite sign.
Also, note that a positive charge at -d will produce a positive field (i.e. pointing to the right), whereas a positiv charge at +d will produce a field pointing the other way.

3. Jan 28, 2014

### collinsmark

As haruspex mentioned, you might want to get rid of the absolute values. They don't really help out much. We can just assume that Q1 and Q2 are the same sign. And since you will eventually be setting the whole thing equal to zero (since you are going to find the location where E = 0), you can even assume that Q1 and Q2 are positive constants without any loss of generality. [Edit: And also as haruspex mentions, although the charges are of the same sign, the electric field from each charge will be in opposite directions. So you'll have to take that into account by putting a negative sign in there somewhere.]

But more importantly, the equation is not correct in terms of the denominators. As it stands now, you have d as the distance in both denominators. That is an impossible situation. The distance value in a given denominator (which gets squared) should be the distance between the charge and some location, x0; not the entire distance between the two charges.

Consider that x0 can be anywhere along the x-axis between -d/2 and d/2. Find the distance between -d/2 and x0, square it, and put that in the denominator under Q1. Similarly, find the distance between +d/2 and x0, square it, and put that under Q2.

Then set E = 0 and solve for x0.

Last edited: Jan 28, 2014
4. Jan 30, 2014

### K.QMUL

Thanks, but how do I use a Cartesian system? I unsure how to do the first part

5. Jan 30, 2014

### collinsmark

That just means to express your distances, positions, and vectors along rectangular, x-y-z-axis system.

Examples of systems that are not Cartesian system are ones that use cylindrical coordinates with ρ-θ-z-axes, or spherical coordinates along r-θ-φ-axes.

Last edited: Jan 30, 2014
6. Jan 30, 2014

### haruspex

As soon as you drew the x and y axes you were using a Cartesian system. (It doesn't say you have to use vectors.)
There is a third problem with your equation that I missed before. On the left hand side you have a vector, on the right scalars. You need to decide whether you want to work with vectors or scalars and stick to that.
E.g. could turn the equation into one concerning the x-component of the field by writing $\vec E . \vec {\hat x}$ on the left (where $\vec {\hat x}$ is the unit vector in the positive x direction). Then, as mentioned before, throw away the modulus signs around the charges. Finally, figure out what the sign should be on each of the two terms inside the bracket: is the field from that term in the positive x direction or the negative x direction?
If you want to write a vector equation then leave the LHS as $\vec E$ but change the terms on the RHS into vectors, like $\frac{Q_1}{d^2}\vec {\hat x}$.