Physics Friday 40

What is the distribution of induced charge on a conducting sphere of radius R placed in a uniform electric field E0?

Let us place the origin at the center of the sphere, and choose the positive z axis in the direction of E0. Now, consider the sphere without the uniform field, and instead, with a charge of +Q placed on the z axis at z=-a and a charge of –Q placed at z=+a, with aR. Then, as noted here, the field near the origin is , and if we take a,Q→∞ with held constant, the result approaches a uniform electric field. Thus, we choose that Q=2πε0E0a2 (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 . Similarly, the +Q charge at z=-a has an image charge of at . The distance between these charges is , and the product of this distance and charge of the image points is
. As a→∞, d→0, and we see from here that the dipole moment p=4πε0R3E0 is held constant in our limit. Thus, in our limit, the image charges approach the point dipole of dipole moment p=4πε0R3E0.

The potential for a uniform field is φ=-E0z. The potential for a point dipole is
, which, in this case, with θ the angle from the positive z-axis (as usual), then , and

Thus the total potential outside the conducting sphere is
And we see by inspection that the potential is constant (specifically, zero) when r=R, the surface of the sphere, as required.

Now, the outward surface normal for the surface of our conductor is just the unit radial vector, as our conductor is a sphere, and so the directional derivative of the potential along that normal is just . Using the results of this post, we see the surface charge density is
which is independent of the radius of the sphere, and which we see integrates over the surface to give a net induced charge of zero.


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