Calculating Electric Fields: 108 Electrons in a Circle

[bneb]

Homework Statement

108 electrons are equally spaced around a circle of radius 10.0 cm.

What is the electric field at the center of the circle?
What would be the electric field be at the same location if one electron was removed?

Homework Equations

\begin{flalign*}E & = & & k \ \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r^2}\hat{r}\\\end{flalign*}

The Attempt at a Solution

For the first part we know that the electric field is zero because for every electron on one side there is exactly one opposite and the E from each cancel out.

The second part is where I am a little confused. We are removing one of the electrons so basically we are making one of them negative? Since before the field was zero really we just need to find the field for the one negative electron. Because it's negative it will change directions to be from removed charge to center of circle? r = 10 cm. What is q and what exactly are we solving for here and how?

 

[rl.bhat]

Hi bneb, welcome to PF.
When you one electron, field due to one electron is left unbalanced. Find the field due to one electron at the center. q is the charge on the electron.

 

[jrlaguna]

You're right, is like having an "anti-electron" added on the initial wheel... In chemistry and solid state physics we call them "holes", and they behave as normal electrons, only with opposite charge... Do you need more hints? :)

 

[nasu]

The field produced by 108 electrons is zero. By superposition, this is the field of one electron + the field of 107 electrons (and is equal to zero).
The field of one electron is easy to calculate.
Then find the field of 107 electrons.

 

[bneb]

Still just learning the basics here so is q = -1, r = .1, and \epsilon_0 = 8.854187817...×10^-12 F/m as per http://en.wikipedia.org/wiki/Electric_constant?