Electric Field and charged plane

1. Mar 12, 2014

Hitchslaps

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

An infinite, charged plane/plate has a uniform positive charge density of σ. Another positively charged particle is found at a distant of D from the plane. In point P, positioned between the two, the electric field equals 0.
A. What is the distance between point P and the charged particle q?
B. The plane is removed and replaced with a new positively charged particle Q. What should be the value of Q in point A in order for the electric field to remain 0?

2. Relevant equations

E=σ/2ε , E=k * Q/r^2

3. The attempt at a solution

In A, I've calculated the electric field the plane exerts on point p, which is E=σ/2ε, and then added the electric field exerted by particle q, which is E=k * Q/(D-r)^2.
E=k * Q/(D-r)^2 + σ/2ε =0

In B, just use this equation E=k * Q/r^2 instead of E=σ/2ε, correct?

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Last edited: Mar 12, 2014
2. Mar 12, 2014

BvU

Hi Hitch, and welcome to PF.

Good use of the template, clear statements and I think your attempt is good. I wouldn't approach it any different -- although: Did you find an imaginary r ? Because everything I see in the expression looks positive...

Tackling B in the way you propose is fine, too.

3. Mar 12, 2014

Hitchslaps

Thanks :)

How would you find the distance from point P to the particle q? I've thought of another way: maybe use a new variable such as x, and then subtract x from D, as in E=k * Q/x^2 ; distance from p= D-x?
I'm unsure.

Last edited: Mar 12, 2014
4. Mar 12, 2014

BvU

You have already found this distance , namely in part A. You called it r.
Now you want to calculate what charge is needed to create the same field at point P as the plate did. Not so difficult!

5. Mar 12, 2014

Hitchslaps

So in B I use E=k * Q/(D-r)^2 with r, the distance I found in A. Thanks again!