Physics Friday 12

Suppose we have a charged, isolated conducting sphere with radius R and total charge Q. Suppose we place a point charge of charge q a distance r>R from the center of the sphere. What is the electrostatic force on the point charge? What is the work done in moving the charge from a distance r1 to a distance r2 on a radial path, assuming the motion is sufficiently slow that we can treat the intermediate states as approximately electrostatic?



We choose our coordinates with the origin at the center of the sphere, and the point charge on the positive z axis.

Note that the point charge alters the charge distribution on the sphere, as the sphere is a conductor. So here we use the method of image charges.

Were we to have a grounded (Φ=0) sphere, we can replace the charge distribution on the sphere with an image charge of charge placed on the positive z axis at a distance from the center of the sphere. The total charge induced on the surface of such a sphere is .

Now, to obtain our sphere of charge Q, we build via linear superposition. We take this grounded conducting sphere and add a uniformly distributed charge Q-q’. This makes the total charge Q. Note, however, that a uniform charge on the surface of a sphere produces a potential and force outside the sphere equivalent to that given if the uniform charge were concentrated at the center of the sphere.

Thus, the force on the point charge due to the charged sphere is equal to that which would be exerted on the point charge due to a charge on the z axis at and a charge Q-q’ at the origin. Using Coulomb’s Law:




Note that when Q and q are of opposite sign, the force is always attractive. When Q and q are the same sign, however, the force is repulsive only when r is greater than some critical value (dependent on R and the ratio of Q and q), which is larger than R; if the point charge gets closer than that, the force becomes attractive. Letting and , then the force is zero when
. This is a quintic equation, and cannot be solved analytically in the general case. However, we see that for , we find that the (unstable) equilibrium point is at . Similarly, for , we find . Lastly, for Q=q, η=1, and we have:



For which the root greater than one is when (which is the golden ratio φ)

The work required to move the point charge radially from r=r1 to r=r2, assuming the motion is sufficiently slow that we c (timescales much longer than the time for light to transit the system) is given by:





Now, note that for , we obtain:
.
We note first that, again assuming that the charges q and Q are of the same sign, the work to remove to infinity is at a minimum when the initial position r1 is the equilibrium position we found earlier. Secondly, we see there is a value where the work changes signs; using η as defined before, and defining , the work to infinity is zero when , or solving the cubic:
.
[With appropriate choices of branch cuts for the (complex) roots when , such that ρ1 is real, and continuous as a function of η at .]
When r1 is less than this critical value, the work becomes positive: it takes more energy to overcome the close-range attraction than is provided by the repulsion in moving from the equilibrium to infinity.

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2 Responses to “Physics Friday 12”

  1. Physics Friday 38 « Twisted One 151’s Weblog Says:

    […] leads to a distribution of charges on the surface of the sphere (”induced charges;” see this previous problem for an example), as required by the resulting discontinuity in the electric field. Note that since […]

  2. Physics Friday 40 « Twisted One 151’s Weblog Says:

    […] (E0=|E0|), then the limit as a→∞ is the desired uniform field. Now, as in this post, we use the method of image charges. The point charge -Q at z=+a has an image charge of at . […]

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