A Potential Problem with a Sphere

Busy busy busy again today so I thought I’d post a quick entry to the cute problems folder. I set this as a problem to my second-year Theoretical Physics students recently, which is appropriate because I encountered it when I was a second-year student at Cambridge many moons ago!


HINT: You can solve this by finding the general solution for the potential at any point inside the sphere, but that isn’t the smart way to do it!

FURTHER HINT: The question asks for the Electric Field at the origin. What terms in the solution for the potential can contribute to this?

Answers through the comments box please!

OUTLINE SOLUTION: A numerically correct answer has now been posted so I’ll give an outline solution. The potential V inside the sphere is governed by Laplace’s equation, the general solution of which is a series expansion in powers of r and Legendre polynomials, i.e. rn Pn(θ). The coefficients of this expansion can be determined for the given boundary conditions (V=V0 at r=a for θ = +1, V=0 for cos θ = -1). However this is a lot more work than necessary. The question asks for the electric field, i.e. the gradient of the potential, and if you look at the form of the potential there is only one term that can possibly contribute to the field at r=0, namely the one involving rP1(cosθ) =rcos θ (which is actually z). Any higher power of r would give a derivative that vanishes at the origin. Hence we just have to determine the coefficient of one term. Using the orthogonality properties of the Legendre polynomials this can easily be seen to be 3V0/4a. The electric field is thus -3V0/4a in the z-direction, i.e. vertically downwards from the top of the sphere.

33 Responses to “A Potential Problem with a Sphere”

  1. Interesting. I have no idea, but I look forward to the solution of how it is done. It has been years since I did electronic engineering and my understanding of maths and electronic principles has since dissipated into thin air.

  2. George S. Williams Says:

    The real hint was with the tags Electrostatics, Laplace’s Equation, and spherical harmonics. Without rechecking the mess I made of the math, I got 3V0/a

    P.S. Did anyone get the right answers to the Magintudes problems last month?

  3. V0/2?

  4. V_0/2?

  5. Rob Old Says:

    Somehow I hardly used Laplace’s equation in my EM courses, so I googled it and found http://mathworld.wolfram.com/LaplacesEquation.html

    … Which leads me to lazily guess without doing any maths that the answer is Vo /2a.

    (If I’m wrong, the maths will have to wait until tomorrow’s lessons are planned!)

  6. Anton Garrett Says:

    By symmetry, the difference in potential between the centre point (at potential P) and one hemisphere (at V_1) must be the same as the difference in potential between the centre point and the other hemisphere (at V_2). So

    P – V_1 = V_2 – P

    from which P is the mean of the two potentials, ie (1/2)V_0.

  7. Anton Garrett Says:

    This has got to be doable by the method of images. As I recall you can simulate a sphere of uniform potential with a couple of well-placed charges of appropriate magnitude, while the plane separating the hemispheres obviously has E perpendicular to it everywhere and is therefore of uniform potential and like an “earth” ie a “mirror”.

  8. This might not be the prettiest solution (or, indeed, correct) but here goes.

    Inside the sphere there is no charge, so Laplace’s equation holds. My the mean value theorem for harmonic functions, the value of V at r is equal to the average over a spherical surface centred at r. So, if we centre at r = 0 (the origin) and consider a sphere of radius a, we find that V(r = 0) = V_0/2.

    Now, suppose the shell has a very small thickness da, and consider a sphere centred just above the origin at z = dz, where dz < da. This sphere will still be inside the shell, but we have shifted a small fraction of its area (equal to a band around its equator of area 2 pi a dz) into the V = V_0 region. Thus, taking the average:

    V(r = dz) = V_0 (2 pi a^2 + 2 pi a dz) / (4 pi a^2) = V_0/2 (1 + dz / a) .

    Now, by symmetry the field at r = 0 only varies in the z direction, so
    E_z = – dV / dz = [V(r = dz) – V(r = 0)] / dz = – V_0 / (2a)

    • On second thoughts, I’m not sure if the sphere over which the average is done is allowed to be inside the shell, since Laplace’s equation doesn’t hold in there (rho \= 0)

  9. Toffeenose Says:

    There is reflection symmetry about the centre horizontal plane, so the potential at the midpoint must be V/2. If the field along the vertical axis were uniform, it would be -gradV=-V/2a. Even if the field is not uniform along the vertical axis, it must also obey the reflection symmetry, and thus have an inflection point at the centre and the value of -gradV at that point must be equal to the average value over the whole axis, namely -V/2a.

  10. How about 3V/(2a)?

  11. Everywhere except on the sphere there are no charges, so that Delta V = 0.

    If you decompose V as a sum of harmonic functions, then, V being regular at r = 0, you have

    V = Sum_{l,m} A_l,m r^l Y_lm (theta, phi).

    By cylindrical symmetry, only the m=0 terms contribute here, so that

    V = Sum_l A_l0 r^l Y_l0 (theta, phi).

    By symmetry again, along the z axis, only the z component of E = -grad V is of interest here, and it reads something like

    E_z = – (cos theta \partial r – sin theta (1/r) \partial theta) V .

    At r = 0, I think that only the l=1 term contributes here, and given that the prefactor of Y_10 is sqrt(3/4pi), what is left is

    E_z (r = 0) = – A_1 sqrt(3/4pi) .

    So all what is left is to compute this A_1.

    It doesn’t give yet the answer, but at least follows the hints.

  12. Themos Tsikas Says:

    It must be zero, then.

  13. domenico Says:

    Jackson give the right solution, with Green function, too complex to write in only 100 written lines (and without latex) only outside the sphere.
    So, I try a smart way: there is a path in the equipotential hemispheres (without electric field), and along the diameter of the sphere; the field of the sphere is symmetric, so that the electric field have the direction of the diameter (that connect the centres of the upper and lower hemisphere), and it is constant because of the Gauss’s law (there is not transversal flux).
    If the distance between the Hemisphere is little, then the charge distribution don’t change with the distance between hemispheres, and the electric field multiplied by the distance tends to zero in the insulated equator.
    I suppose that the field along the diameter is V_0/(2a), because the line integral of the electric field is zero along a closed path.
    If it so, then the problem use all the electrostatic laws.

  14. Toffeenose Says:

    Zero! Close to the top, it approximates a complete metal sphere at potential V, and close to the bottom a metal sphere at potential zero. Thus the line of equipotential is vertical at those two points, but of value V at the top and of value 0 at the bottom. The potential at the mid point is V/2 by symmetry, and the plot of potential along the axis must have mirror symmetry. For all this to be valid, the line of equipotential must cross the centre vertically – which means the electric field is zero.

    • telescoper Says:

      No. That’s a very wrong argument!. In fact in the neighbourhood of the origin the equipotential is in the equatorial plane, perpendicular to the direction you claim. By symmetry the electric field runs vertically through the origin, which tells you that the potential must be proportional to z there.

  15. Navneeth Says:

    In honour of Adams’ birth anniversary: 42.

  16. Assuming that the insulator is infinitely thin (big assumption I know), and the sphere is a perfect conductor, then the potential on the surface of the sphere will have a step function at the equator. I wonder if there might be a step function even inside the sphere. In which case the answer would be minus infinity.

    Although it makes more sense that Luke Barnes is right. You didn’t post “nope” after that comment.

  17. Bryn Jones Says:

    Well, I didn’t think of using spherical harmonics. Therefore the only way I could think of doing it involved integration over solid angles, but didn’t get around to doing it. I didn’t see any shortcuts.

  18. Themos Tsikas Says:

    The equivalent example is worked out in full detail in Mary Boas’s book Mathematical Methods in the Physical Sciences, page 667, STEADY-STATE TEMPERATURE IN A SPHERE. Nice problem.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: