The Electric field between two charges

[K.QMUL]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[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.

 

[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.

 

[K.QMUL]

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

 

[collinsmark]

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

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.

 

[haruspex]

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

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 $\frac{Q_1}{d^2}\vec {\hat x}$.