In summary, Homework Statement discusses two concentric spherical conducting shells. The inner sphere has radius r1, potential V1, while the outer sphere has radius r2, potential V2. The potential at the center of these two sphere, at r0=(r1+r2)/2, is found to be double the correct answer.
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Vibhor
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Is ##\frac{V_2r_2^2-V_1r_1^2}{r_2^2-r_1^2}## the answer ?
Ok, since the cat is out of the box, let us reconnect to the physics and the more physical arguments way of doing the problem and see how it relates to the "shut-up-and-solve-the-ODE" approach.
If we can find a charge configuration in space which does not contain any charge between the shells and reproduces the correct potentials on the shells at ##r_1## and ##r_2##, then the corresponding potential for points between the spheres will be the solution to the problem.
Since the problem is obviously spherically symmetric, we are only interested in spherically symmetric solutions with no charge between the spheres. Between the spheres, any spherically symmetric charge distribution at ##r < r_1## is going to produce a potential proportional to ##1/r## while any spherically symmetric charge distribution at ##r>r_2## is going to produce a constant potential according to the shell theorem (it is also possible to ignore the charge outside and just refer to potentials being defined up to a constant addition). Thus, the most general spherically symmetric solution between the spheres is given by
$$
V(r) = \frac Ar + B,
$$
with ##A## and ##B## being constants given by the charges.
Of course, this is exactly the same result that was obtained by solving the differential equation and the determination of ##A## and ##B## (and thus, the charges) is performed in the same way.
If you want an even more "Physical" solution, without differential equations, determine the electric field between the shells. Using symmetry again, we assume that E is radial and its magnitude is equal in any direction. Applying Gauss' Law at distance r from the centre,
[tex]\oint {EdA}=4πr^2E=Q_1/ε_0 [/tex]where Q1 is the charge on the inner sphere (unknown yet) so ##E=\frac{Q_1}{4\piε_0}\frac{1}{r^2}=kQ_1\frac{1}{r^2}##
The electric field is negative gradient of the potential, U. ##\vec E(r)=-\nabla U(r)##.
The potential at a point between r1 and r2 is obtained by integrating the equation above between r1 and r: