Charges placed at corners of a square

1. Feb 12, 2010

krtica

What would be the general approach to the problem? This isn't a homework question..

Q: Charges of +4.8 are placed on corners of square, each side being 8m. What is the electric potential at the center of the square?

2. Feb 12, 2010

CompuChip

The potential satisfies a superposition principle. So V(r) = V1(r) + V2(r) + ... + Vn(r), where Vi(r) is the position at position r due to the i-th charge.

In this case, you can use the symmetry, so at the origin
V(0) = 4 V1(0)
with V1(0) the potential at the origin due to a charge on one corner of the square.

Note that the electric field at the origin will be zero, as the origin is an equilibrium point: all the charges are attracting or repelling your test charge with the same force, so it will stay put.

3. Feb 12, 2010

krtica

Thank you. Does that imply that the center of the square would have zero potential?

4. Feb 12, 2010

ideasrule

No. Do the calculation. Figure out the electric potential induced by each charge.

5. Feb 13, 2010

CompuChip

Recall that the force on the charge is F = q E, with q the charge and E the electric field, while E = -grad V, with V the potential.
So it only implies that V is at a local minimum or maximum (or saddle point) such that grad V = 0, it doesn't say anything about the value of V. (Of course, if you want, you can make it 0 by shifting the potential everywhere by some constant. The reason is that only potential differences have physical meaning, so we can always choose the value of the potential at one point. Usually we let it be zero at some point infinitely far away, though, from which the textbook formula for the potential of a point charge derives).