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GRE question for the week of 10/8/09 - Oct 8, 2009 at 10:08AM
Dave Avatar
Dave
San Marcos, Texas
413 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...


Adjunct Professor of Physics, Editor of The Physics Teacher, and GWU SPS Chapter Advisor


Re: Re: GRE question for the week of 10/8/09 - Oct 15 2009 5:45PM
Dave Avatar
Dave
San Marcos, Texas
413 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


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