Electrostatic potential and electric fields

[secret5437]

Homework Statement

3. Four charged particles are held fixed at the corners of a square of
side s. All the charges have the same magnitude Q, but two are
positive and two are negative. In Arrangement 1, shown ,
charges of the same sign are at opposite corners. Express your
answers to parts (a) and (b) in terms of the given quantities and
fundamental constants.

(a) For Arrangement 1, determine the following.
i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square

The bottom two charged particles are now switched to form Arrangement 2,
shown , in which the positively charged particles are on the left and the
negatively charged particles are on the right.

(b) For Arrangement 2, determine the following.

i. The electrostatic potential at the center of the square
ii. The magnitude of the electric field at the center of the square

In which of the two arrangements would more work be required to remove the particle at the upper right corner from its present position to a distance a long way away from the arrangement? Justify your answer.

Homework Equations

V = KQ/r
E = KQ/r²

The Attempt at a Solution

I'm pretty sure the answer for Arrangement 1 is 0, for parts i. and ii.
Help on Arrangement 2, please?

 

[Delphi51]

You haven't given any information on what arrangement 2 is.
Or on what you did for part (a). We helpers don't really like to work out all the details to check an answer - that's your job! We love to help with the understanding, though.

 

[secret5437]

I know that the answer to Arrangement 1 is 0, because the two positive and two negative charges cancel each other out.

 

[secret5437]

and arrangement 2 looks like this:

Q+......Q-




Q+......Q-






Arrangement 1 looks like this:
Q+.....Q-





Q-......Q+


(imagine those two arrangements as squares)

 

[Delphi51]

Oh, clever of you! Yes, I think you are right, though it seems a bit surprising at first that you can just add the potentials due to each charge. I looked it up to make sure!
I guess that is the clue for the second arrangement!