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GR9677 #3
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Electromagnetism$\Rightarrow$}Potential

Recall the elementary equations, $V = \int \vec{E}\cdot d\vec{l} = \int \frac{dq}{4\pi \epsilon_0 r}$.

$r=\sqrt{R^2+x^2}$, and $dQ=Q$ thus $V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}$, as in choice (B).

Alternate Solutions
 jmason862009-09-28 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 $V\prop \frac{1}{r}$ so just get r. Draw the triangle and see that (B) does the job. Reply to this comment spacebabe472007-09-19 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. Reply to this comment sblusk2006-10-24 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.Reply to this comment
fredluis
2019-08-08 06:54:37
After the binomial expansion, the potential should have a \"-\" instead of a \"+\" in the expansion term. However, a minus exists in the force equation because the test charge has charge -q. tile contractor
sean9
2017-11-28 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
2017-10-25 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 non-zero 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
2017-10-16 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
2009-09-28 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 $V\prop \frac{1}{r}$ so just get r.

Draw the triangle and see that (B) does the job.
spacebabe47
2007-09-19 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.
 ewhite22007-10-30 17:32:36 how is C eliminated?
 kyros2007-11-01 10:40:09 C isn't eliminated that way -but it's clear that at x = 0, the answer should be non-zero(consider a positive charge and positive Q, the integral from infinity to zero can't be zero.)
 ghost2k152017-08-27 01:13:21 How does that eliminate C?
 ghost2k152017-08-27 01:14:12 sorry was on a half screen and didn\'t see the comments
sblusk
2006-10-24 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.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$