GR9677 #3



Alternate Solutions 
jmason86 20090928 18:37:55  Don't even both with anything more than the distance. All the answers are identical except for the dependence on R and x which are the distances. You know that so just get r.
Draw the triangle and see that (B) does the job.   spacebabe47 20070919 13:39:58  As R > 0, the equation should simplify to the potential for a point charge. This eliminates C, D and E. The equation should also have some R dependence, this eliminates A. Choose B.   sblusk 20061024 06:14:40  An alternate way would be to just use directly:
V(r) = Integral (1/4*(pi*e_0))* dQ/r
where r=sqrt(R^2+x^2).
This is in principle easier because one is dealign with a scalar, and not the electric field which is a vector.  

Comments 
sean9 20171128 08:41:44  When x = 0, the answer is just that of a point charge Q at distance R.\r\njson formatter\r\n   curious_onlooker 20171025 19:59:53  When x = 0, the answer is just that of a point charge Q at distance R. You can see this at a glance because the integrand (if you wrote it out) would not have any position dependence (all dq\'s are the same distance from where we\'re evaluating the potential). When x is nonzero but R = 0, the answer should also be that of a point charge but with distance x. The only answer satisfying this is B.   Laxrabi23 20171016 12:58:32  I thought I could get to see and discuss all released GRE questions in here, but I can only see discussions about the four old question papers. Are only those discussed here or am I not able to find discussions about recently asked questions? Please someone make me clear about it. Thanks,,\r\n   jmason86 20090928 18:37:55  Don't even both with anything more than the distance. All the answers are identical except for the dependence on R and x which are the distances. You know that so just get r.
Draw the triangle and see that (B) does the job.   spacebabe47 20070919 13:39:58  As R > 0, the equation should simplify to the potential for a point charge. This eliminates C, D and E. The equation should also have some R dependence, this eliminates A. Choose B.
ewhite2 20071030 17:32:36 
how is C eliminated?

kyros 20071101 10:40:09 
C isn't eliminated that way but it's clear that at x = 0, the answer should be nonzero(consider a positive charge and positive Q, the integral from infinity to zero can't be zero.)

ghost2k15 20170827 01:13:21 
How does that eliminate C?

ghost2k15 20170827 01:14:12 
sorry was on a half screen and didn\'t see the comments

  sblusk 20061024 06:14:40  An alternate way would be to just use directly:
V(r) = Integral (1/4*(pi*e_0))* dQ/r
where r=sqrt(R^2+x^2).
This is in principle easier because one is dealign with a scalar, and not the electric field which is a vector.  

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