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 a≫R. 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.

Suppose we have a group of n people (n≥2). The name of each person is written on a separate slip of paper. These n slips are put in a box, and each person then draws a slip from the box. What, then, is the probability that nobody draws their own name? What happens to this probability when n becomes large?


We can treat a drawing outcome as a permutation on n items (the names). There are n! such permutations. What we need to find the probability we desire is the number of these permutations which leave no elements unshifted; that is, which contain no fixed points.

For the case n=2, there’s one such permutation (the swap), and so the probability is . Similarly, for n=3, there are the two cyclic permutations, and so the probability is . Trying to count the permutations becomes more lengthy with each increase in n. Thus, we seek a simpler way to do this.

Let us consider the probability of having k of the n people draw names other than their own, and write this . Then we seek . Now, , the probability that everyone draws their own name, is . We also have, from the basics of probability, that .

Thus, we see

Now, is the probability of k people drawing the names of others, and thus n-k drawing their own. As these are independent events, we see
(to make this product accurate for k=0, we need ).
Combining this with the basic probability rule, we get

for any n. This gives us the recursive formula

for in terms of all previous ones.
Thus, we can determine the values:
.
Similarly, .

Now, let us examine :






Examining these, we thus find our desired result:
.
(A proof by induction can confirm this.)

And for large n, we use . This tells us , and thus, we see that approaches as n goes to infinity, and the approximation becomes very close very rapidly: .

Let us consider two point charges of charge -q and q (q>0) separated by a distance d. Let d be the vector from the negative to the positive charge. Choosing the point halfway between the charges as the origin of our coordinate system, then the electric potential is given by:

and the electric field is
.

First, let us examine the field far from the charges (r≫d). To simplify things, let θ denote the angle between d and r. Then the law of cosines tells us that
and
. Thus, for r≫d, we see

and

So we have
.
Using , we see that for distant points, the field is approximated by
.
Defining , and using the unit vector in the direction of r , we have
, or, in terms of θ,
, so we see the magnitude of the distant field goes as the inverse cube of distance. The vector p is the dipole moment, and if one lets d go to zero while holding p constant, the field approaches the one above at all points, and one has a point dipole.

Now, let us instead consider the potential and field near the point between the charges (r≪d). In this case, , and

Thus

and so
, where is the unit vector in the direction of d. Note that this approximation is a constant vector. Thus, if we have while holding constant, our electric field appoaches a uniform electric field.

Find .

The methods I’ve shown before for definite integrals (differentiating under the integral sign, converting to a double integral and reversing order, expanding in infinite series) will not be of much use on this one. However, we can use complex analysis and contour integration to find this.

First, due to symmetry, we see . Further, symmetry also tells us , as the integrand is odd. Thus:
.

So we will compute the integral of over an appropriate contour in the complex plane. As , the function has poles at ±i. Expanding in partial fractions, , and so we see the residues of at z=i and z=-i are and , respectively.

Let us first choose the (counter-clockwise) contour bounding a semicircle of radius R>1 in the upper half plane; that is, the contour consists of the real axis from -R to R, and then the semicircle of radius R in the upper half plane. We can parametrize the semicircle as , 0≤t≤1.

The contour encloses only the pole at z=i, so therefore the integral of our function over our contour is given by the residue theorem as:


Breaking up our contour into its two segments, though, we see:
, where C₁ is our semicircle.
Thus

Now, we will need to use the following theorem of complex analysis:
Suppose f(z) is continuous on the contour C, and there exists a real constant M such that on C. Then , where L is the length of the contour C.

So, examining |f(z)| on C₁, we see:
.
For a>0, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1, and thus
.
Since the length of our semicircle is πR, our theorem tells us
. Now, as R goes to infinity, goes to zero. Thus, .

Applying this limit to our equation for the real axis portion,

So, for a>0,
.

Note that if a<0, our bound on the semicircle does not work. So now, let us perform the integral over the contour bounding a semicircle in the lower half-plane; it consists of the real axis from R to -R, and then the semicircle of radius R in the lower half plane, which we can parametrize as , 0≤t≤1.

This contour encloses the pole at z=-i only, so via the residue theorem,


Breaking up our contour into its two segments, though, we see:
, where C₂ is our semicircle.
Thus .

As to |f(z)| on C₂, we see:
.
Here, the numerator in the above is less than or equal to one for all t in the range 0≤t≤1 when a<0, and thus
.
The length of our semicircle is still πR, and so applying the theorem and taking the limit as R goes to infinity, we get
.
Therefore , and so for a<0.

Combining these two, we see:
.

(Note that a=0 gives the correct equation ).

I was looking through some older bookmarks (specifically, I was looking for some pages for a multi-part work that’s been percolating in my mind for months, and which I’m now putting to (metaphorical) paper), and I found these two posts that describe the “Care Bear Stare” mode of political thought found in segments of the American political spectrum, both Left and Right. Namely, this is the belief that all problems are surmountable if only we care hard enough (Left), or have enough will (Right).
“Care Bear Stare!” by Julian Sanchez
“The Care Bear Stare Won’t Fix the World” by Rod Dreher

Electrostatics and Charged Surfaces

In the integral form, Gauss’ Law reads:

or in words, that the flux of an electric field through a closed surface equals the total enclosed electric charge divided by ε₀. Now, let us consider some surface with a surface charge density σ(r). Let us pick a point on the surface. Let us call the two sides of the surface side 1 and side 2, and let n be a surface normal at the point, directed from side 1 to side 2. Lastly, let us call the electric field immediately on either side of the surface E₁ and E₂.

Now, let us make a gaussian surface around our point. Our gaussian “pillbox” will be a small, flat cylinder with ends of area dA and axis parallel to n. Letting the height of the cylinder be small, the flux through the “sides” will vanish, and the net flux will be that through the ends. For the end on side 2, we see the (outward) flux will be, for a small “pillbox”, E₂·(dAn)=dAE₂·n. Similarly, the flux for the end on side 1 is E₁·(-dAn)=-dAE₁·n. Thus, the total flux is (E₂-E₁)·n dA. Now, the total enclosed charge is just σ dA, and so:
.
Thus, a surface charge creates a discontinuity in the field perpendicular to the charged surface.

Now, let us consider a perfect conductor in an electric field (the basis of many electrostatics problems). The requirement for a static situation is that the electric potential be constant on the surface of the conductor, and that the electric field be zero within the conductor (otherwise charges in the conductor will be experiencing net electric forces, and thus no static state). As a result, this generally 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 the definition of the electric potential Φ is that E=-∇Φ, the condition that Φ is constant on the surface of the conductor is equivalent to the requirement that the electric field be normal to the surface.

Now, combining these conditions with our finding about surface charges, we see that at the surface of the conductor, with our conventions as before, and side 1 the interior of the conductor, E₁=0, and so σ= ε₀E₂·n. Writing this in terms of the potential outside the conductor, we have σ=-ε₀∇Φ·n=-ε₀∇_nΦ; where ∇_nΦ=∇Φ·n is the directional derivative in the direction of the outward surface normal n. This allows us to find the induced charge on the surface of a conductor from the electric potential outside the conductor.

The United States Constitution was adopted by the Constitutional Convention in Philadelphia 221 years ago today. [My thanks go to Eugene Volokh for the reminder].

When written as a decimal, the fraction 1/89 consists of a repeating decimal; specifically with a 44-digit repeat:
1/89=0.01123595505617977528089887640449438202247191…
Note the first few digits after the decimal point: 0, 1, 1, 2, 3, 5. You should recognise from last week the start of the Fibonacci sequence. Examing this, we see:

Or, using our convention last week, it looks like the series

is equal to 1/89. Does the sum actually hold? What if we try it in bases other than base 10?


In general, with base b (b≥2, of course), our sum in this fashion is:

Now, recall that last week we found the formula
,
where is the golden ratio.
This means our sum is:
.

Note that both sums in the result above are convergent geometric series, and as , we see that
,

and for b=10, , so the original infinite sum is 1/89. We see that for binary (b=2), we have the sum
.

Note that the zeroes of the quadratic polynomial are and , which are the terms raised to the power of n in our formula for F_n.

Fifty years ago today, Jack Kilby at Texas Instruments demonstrated the first integrated circuit.

Let us consider a pair of parallel, cylidrical metal rails of radius r, the centers of which are separated by a distance d. A movable metal connector is placed between these, so as to slide freely along the rails.


Now, suppose we run a current I through this by connecting a current source to the rails at one end:


What will be the resulting force, if any, on the connector?


Assuming the length of the wires is at least several times d, we can use the “infinite wire” formula for the magnetic field resulting from the current in a rail. At a distance R>r from the center of the wire, the field is , in the direction given by the right-hand rule.


Thus, the field at a point along the connector a distance x from the center of one rail will be


Now, the magnetic force on a current-carrying wire of length L, sometimes called the Laplace force, is given by:
F=LI×B.
Here, our connector, and the current in it, are perpendicular to the magnetic field; the cross product I×B has magnitude IB and is directed away from the end to which the current source is connected.

Thus, for an infinitesimal segment dx of our connector, the force on the segment is

To find the total force, we integrate this over x from r to d-r:


So the connecter will experience a force seeking to propel it along the rails. This physics is the basic principle of the rail gun. Since μ₀=4π×10^-7 N·A^-2, the non-geometric term will be . Thus, for d/r~10, to obtain forces on the order of a newton, the current needs to be on the order of a kiloampere. Thus lies the difficulty of the railgun; massive currents and voltages are required to obtain effective projectile launching.