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Lisa...
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The question 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?
The first remark
: C= \frac{Q}{V}
In this case Q= Q = the absolute charge on every sphere.
V= the potential difference between #1 and #2, given by V= - \int_{#1}^{#2} E dl
with dl= dxi +dyj+dzk
The electric field due both spheres is basically Coulombs law (E= \frac{Q}{4 \pi \epsilon_0 r^2} with the only difference is for #1 Q= Q and for #2 Q= -Q. Between the two spheres the two electric fields add (if you draw a pic, the field lines of #1 move away from #1 (cause it's positive) in radial direction, and the field lines of #2 approach #2 in radial direction (cause it's negative)).
Therefore E total= \frac{2Q}{4 \pi \epsilon_0 r^2}.
Now substitute this in the integral V= - \int_{#1}^{#2} E dl with values of:
E= E total= \frac{2Q}{4 \pi \epsilon_0 r^2}
dl= dr, with dr the projection of dl in radial direction and
#1= -0.5 l
#2= 0.5 l
(I placed the origin exactly in the middle of both spheres, so when integrating the positions of #1 (at -0.5 l)and #2 (at 0.5 l) I need to integrate between -0.5 l and 0.5 l.
This gives:
V= - \int_{-0.5 l}^{0.5 l} \frac{2Q}{4 \pi \epsilon_0 r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \int_{-0.5 l}^{0.5 l} \frac{1}{r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \left[ - \frac{1}{r} \right]_{-0.5 l}^{0.5 l} =<br /> - \frac{2Q}{4 \pi \epsilon_0} \frac{-4}{l}= \frac{8Q}{4 \pi \epsilon_0 l} = \frac{2Q}{\pi \epsilon_0 l}
Substituting this into the formula of C= \frac{Q}{V} provides:
C= \frac{Q}{\frac{2Q}{\pi \epsilon_0 l}}= \frac{Q \pi \epsilon_0 l }{2Q}= \frac{\pi \epsilon_0 l }{2}
So I am I right or not ...?
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?
The first remark
: C= \frac{Q}{V}In this case Q= Q = the absolute charge on every sphere.
V= the potential difference between #1 and #2, given by V= - \int_{#1}^{#2} E dl
with dl= dxi +dyj+dzk
The electric field due both spheres is basically Coulombs law (E= \frac{Q}{4 \pi \epsilon_0 r^2} with the only difference is for #1 Q= Q and for #2 Q= -Q. Between the two spheres the two electric fields add (if you draw a pic, the field lines of #1 move away from #1 (cause it's positive) in radial direction, and the field lines of #2 approach #2 in radial direction (cause it's negative)).
Therefore E total= \frac{2Q}{4 \pi \epsilon_0 r^2}.
Now substitute this in the integral V= - \int_{#1}^{#2} E dl with values of:
E= E total= \frac{2Q}{4 \pi \epsilon_0 r^2}
dl= dr, with dr the projection of dl in radial direction and
#1= -0.5 l
#2= 0.5 l
(I placed the origin exactly in the middle of both spheres, so when integrating the positions of #1 (at -0.5 l)and #2 (at 0.5 l) I need to integrate between -0.5 l and 0.5 l.
This gives:
V= - \int_{-0.5 l}^{0.5 l} \frac{2Q}{4 \pi \epsilon_0 r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \int_{-0.5 l}^{0.5 l} \frac{1}{r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \left[ - \frac{1}{r} \right]_{-0.5 l}^{0.5 l} =<br /> - \frac{2Q}{4 \pi \epsilon_0} \frac{-4}{l}= \frac{8Q}{4 \pi \epsilon_0 l} = \frac{2Q}{\pi \epsilon_0 l}
Substituting this into the formula of C= \frac{Q}{V} provides:
C= \frac{Q}{\frac{2Q}{\pi \epsilon_0 l}}= \frac{Q \pi \epsilon_0 l }{2Q}= \frac{\pi \epsilon_0 l }{2}
So I am I right or not ...?
!