How to Calculate Electric Potential Energy in an Infinite Grid of Charges?

AI Thread Summary
The discussion focuses on calculating the electric potential energy of a cation in an infinite grid of alternating anions and cations. The user initially applies the formula for potential energy between two charges but struggles with extending it to an infinite grid. They consider the symmetry of the grid, suggesting that the contributions from charges on either side might cancel each other out. The user arrives at a summation equation but questions its convergence due to the alternating series. Ultimately, they express that they have resolved their confusion, indicating a successful understanding of the concept.
Icheb
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Following exercise:
There is a one dimensional grid of infinite size and it consists of anions and cations (each with a charge of 1e / -1e) like this:

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The distance between the anions and cations is 5*10^-10m.
Now I am supposed to calculate the electrial potential energy of one cation alone. I know that, if I only have two charges, the energy would come from

W = 1/(4\pi \epsilon_0) * (q_1 * q_2)/r

But I don't understand how to calculate the energy for a grid with an infinite amount of charges. Can someone point me in the right direction please? I'm not asking for a solution, just for a small hint which will guide me in the correct direction.
 
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not too sure, but if they are alternating then the electric potential energy of one of the "particles" on the left will be equal in magnitude, but opposite in direction to the one on the right, so summing the electric potential energies for all the particles, would give you zero overall... i think, that would be my logic anyway.
 
I don't think it's that easy, but thanks for the answer. :)
 
I have arrived at the following equation now:

\phi = k * \sum_{n=1}^{\inf} 1/n * (-1)^{n+1} * q/r

However, this doesn't seem to make sense to me since -1^n+1 doesn't converge. Did I make an error in my calculations or am I missing something?
 
Never mind, figured it out. :)
 
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