Need help with "charges" question please!!

please help! this must be taught tomorrow in class, and this is my first year teaching. How can I explain it to my students?

Q: What point charges, all having the same magnitude, would you place at the corners of a square (one charge per corner), so that both the electric field and the electric potential (assuming a zero reference value at infinity) are zero at the center of the square? Account for the fact that the charge distribution gives rise to both a zero field and a zero potetial. Explain Thoroughly.

A: ???

## Homework Statement

please help! this must be taught tomorrow in class, and this is my first year teaching. How can I explain it to my students?

Q: What point charges, all having the same magnitude, would you place at the corners of a square (one charge per corner), so that both the electric field and the electric potential (assuming a zero reference value at infinity) are zero at the center of the square? Account for the fact that the charge distribution gives rise to both a zero field and a zero potetial. Explain Thoroughly.

A: ???

no eqn needed..

## The Attempt at a Solution

I assume that there would be a positive charge in top left, negative charge top right, postive charge bottom right, and negative charge bottom left..but I'm sad to say I'm not sure.

Staff Emeritus
Gold Member
So, you have a degree in Physics but cannot explain a simple array of charges?

Staff Emeritus
Gold Member
So, you have a degree in Physics but cannot explain a simple array of charges?

Note to Mentors; identical thread in Intro Physics

I'm sorry, I didn't know not to post twice. I just need help. I've taught Chemistry for 27 years, this year the physics teacher left. So now I'm having to balance both. Can you please help?

arunma
Well we know that electric potential is given by $$V = \frac{kq}{r}$$. As you go around the corners of the square, if you alternate between charges of +q and -q, that ought to do it. This way, the two positive charges will give a potential of $$\frac{2kq}{r}$$, and the two negative charges will give $$-\frac{2kq}{r}$$. Since potentials add as scalars, the net potential will be 0. As for the electric field, the superposition principle makes it fairly obvious that the field will be zero as well.

Anyway, someone let me know if I made a careless error. But I think this should do it.

Gold Member
Well we know that electric potential is given by $$V = \frac{kq}{r}$$. As you go around the corners of the square, if you alternate between charges of +q and -q, that ought to do it. This way, the two positive charges will give a potential of $$\frac{2kq}{r}$$, and the two negative charges will give $$-\frac{2kq}{r}$$. Since potentials add as scalars, the net potential will be 0. As for the electric field, the superposition principle makes it fairly obvious that the field will be zero as well.

Anyway, someone let me know if I made a careless error. But I think this should do it.

Seems like you attempted to give a direct answer to a homework problem. btw, how did $$\frac{1}{r^2}$$ become $$\frac{1}{r}$$ ?

rbj
how did $$\frac{1}{r^2}$$ become $$\frac{1}{r}$$ ?

potential energy is not force.

cowshrptrn
you mean electric field, not force since its just one charge, not 2

Gold Member
you mean electric field, not force since its just one charge, not 2

No, hes right. With the case of two charges, the electric potential will be negative if the charges have opposite sign and positive if the charges have the same sign. Therefore we can see how the energy can be lost by opposing charges, thus satisfying arunma's explanation.

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Staff Emeritus
I've merged the two threads that was cross-posted, so if it doesn't make any sense, it's not my fault.

Seems like you attempted to give a direct answer to a homework problem. btw, how did $$\frac{1}{r^2}$$ become $$\frac{1}{r}$$ ?
I used 1/r because we're discussing potentials rather than fields. Electric potential has one more dimension of length than electric field (also one can remember that electric field has SI units of volts/meter). And with respect to infinity, any single electric charge creates a potential of $$\frac{kq}{r}$$.