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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