(adsbygoogle = window.adsbygoogle || []).push({}); Follow up on the 2 spheres capacitor problem. Help needed before tomorrow!!!

It was probably a bit hard to see what the old topic was about, so I started a new one. I'm still not quite sure on this problem, so I'd REALLY appreciate it if somebody could clear things up a little bit more. The problem is the following one:

Two isolated conducting spheres of equal radius have charges +Q (#1)and -Q (#2) respectively. If they are separated by a large distance compared to their radius, what is the capacitance of this unusual capacitor?

I've drawn this picture:

http://img147.imageshack.us/img147/7725/naamloos6ha.gif [Broken]

The origin is in the center of the left sphere.

This is what I've found till now:

[tex]C= \frac{Q}{V}[/tex]

In this case Q= Q = the absolute charge on every sphere.

V= the potential difference between #1 and #2, given by [tex]V= - \int_{#1}^{#2} E dl = = \int_{#2}^{#1} E dl[/tex]

with dl= dxi +dyj+dzk

The electric field due both spheres is determined by Coulombs law ([tex] E= \frac{kQ}{r^2}[/tex]).

Pick a point at a distance r from the origin somewhere on the drawn x-axis. In that point the E field of #1 and #2 is felt in the positive x direction (#1 is positive, so the field lines take off from #1 and #2 is negative, so the field lines approach #2). Therefore the magnitudes of the electric fields of #1 and #2 add and the resultant E is in the positive x direction.

The distance from the point P to the center of #1 (in the origin)= r.

Therefore [tex] E_1= \frac{kQ}{r^2}[/tex]

The distance from the point P to the center of #2 = [L+2R]-r.

Therefore [tex] E_2= \frac{kQ}{([L+2R]-r)^2}[/tex]

And [tex] E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]

Now substitute this in the integral [tex]V= \int_{#2}^{#1} E dl [/tex] with values of:

[tex] E= E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]

dl= dr, with dr the projection of dl in radial direction and

#1= R

#2= L+R

(that is the domain of the space between the two spheres where the point P can be chosen in)

This gives:

[tex]V= \int_{L+R}^{R} \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2} dr = \left[ - \frac{kQ}{r} + \frac{kQ}{([L+2R]-r)}\right]_{L+R}^{R} =

( - \frac{kQ}{R} + \frac{kQ}{(L+R)}) - (-\frac{kQ}{L+R} + \frac{kQ}{(R)}) =[/tex]

[tex]\frac{2kQ}{L+R} - \frac{2kQ}{R}= \frac{2kQR-2kQ(L+R)}{R (L+R)}= \frac{2kQR-2kQL -2kQR)}{R (L+R)}= \frac{-2kQL}{R (L+R)}[/tex]

Substituting this into the formula of [tex]C= \frac{Q}{V}[/tex] provides:

[tex]C= \frac{Q}{ \frac{-2kQL}{R (L+R)}}= \frac{Q R (L+R)}{-2kQ}= \frac{R (L+R)}{-2k}[/tex]

But given is R<<L, so R= 0 which gives me a C of 0, which can't be possible!!! So what have I done wrong?!

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# Follow up on the 2 spheres capacitor problem. Help needed before tomorrow

**Physics Forums | Science Articles, Homework Help, Discussion**