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I am tasked with the following exercise:

Electric charges of $+q$ are fixed at the four corners of a square of side-length a. (i) What is the electric potential at the centre of this square? (ii) If two of the charges are replaced by charges of $-q$, what now is the potential at the centre of the square? (iii) Does it matter for the electric potential which two charges are replaced? (iv) Does it matter for the electric field at the centre which two are replaced?

And I think this exercise requires me to understand how electric potential contributions are made with varying charge signatures.

For (i), I think the electric potential at the centre of the square should be maximum, as none of the electric potential contributions of each the charges should interfere with eachother. However, for (ii), the fact that potential here can switch sign challenges my knowledge further, and I can't figure out what will happen.

Need some guidance here.

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The electrostatic potential at the center of the square is given by the superposition of the electrostatic potentials of the charges at the four corners which all have the distance $r=d/2$ from the center where $d$ is the diagonal of the square. Thus for for positive charges $=q$ at the corners, the potential at the center is $$\Phi=\frac {4q}{4\pi \epsilon_0 r} \tag 1$$ In the case that there are two positive and 2 negative charges the potential is $$\Phi=0$$ irrespective where they are located on the corners. It does matter for the electrical field in the center where the charges are are located, because the electrical field is a vector, the negative gradient of the potential.

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  • $\begingroup$ I see. Just for my conceptual understanding, if we were to, say, consider $4$ planets at the corners of the square, would this be the same in scenario in terms of gravitational potential? Since there is no concept of charge in gravitational potential, I would be correct in saying the potentials would add up collaboratively? $\endgroup$ – sangstar Feb 26 '18 at 1:26
  • $\begingroup$ @sangstar - The gravitational potential is always negative so that the force between two masses is always attractive. Also for the gravitational potential you can always add the potential of different masses at different locations at a given point to obtain the total potential just like in the case of the charges. $\endgroup$ – freecharly Feb 26 '18 at 2:20

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