## GRE Question of the Week post and replies

GRE question for the week of 10/8/09 - Oct 8, 2009 at 10:08AM

Dave
San Marcos, Texas
352 Posts

Hi Everyone,

I just posted the answer to last week's question, so here is another one for you to think about.

Five positive charges of magnitude q are arranged symmetrically around the circumference of a circle of radius r.   What is the magnitude of the electric field at the center of the circle?  (k=(4*pi*epsilon0)^-1)

A. 0

B. k*q/r^2

C. 5*k*q/r^2

D. (k*q/r^2)*(cos(2*pi/5))

E. (5*k*q/r^2)*(cos(2*pi/5))

I will post the answer next week.

Dave

Try not to become a man of success, but rather try to become a man of value -- Albert Einstein

### Replies to GRE question for the week of 10/8/09

Re: GRE question for the week of 10/8/09 - Oct 15 2009 4:53PM
Gary
Society of Physics...
293 Posts

I say zero for the electric field, since it is a vector, because of symmetry...

but what about the potential---it's not zero, is it (I'm feeeling a little rusty at this...)?...No, otherwise the potential at the center of the positive end of a battery would be zero, also, instead of 1.5 Volts...

Oh, right, the potential is just 5 times that of a single charge at distance r, or 5kq/r, if you take the potential to be zero at infinity...

NSF Program Director (on assignment from the AIP and the Society of Physics Students to serve as the Robert Noyce Scholarship Program Director at the National Science Foundation)

Re: Re: GRE question for the week of 10/8/09 - Oct 15 2009 5:45PM

Dave
San Marcos, Texas
352 Posts

Gary is right on both counts.  A good visualization is given in the attached file.  No matter how many charges there are, as long as they're arranged symmetrically, their field vectors form a regular polygon, and so sum to zero.

Dave

Attached File: GRE charges.pdf

Try not to become a man of success, but rather try to become a man of value -- Albert Einstein