From Gauss theorem, we can show that the surface of a curved charged conductor, the normal derivative of the electric field is given by

where and are the principal radii of curvature of the surface.  Gauss’s law in integral form is expressed as

when there are no charges enclosed in the surface S .

Before considering the three dimensional problem,  we first tackle the problem in two dimensions. We take a curved Gaussian box next to the surface of a  charged conductor at a point where the radius of curvature is R.  Application of Gauss’s law yields

      (2)

where and  are the areas of the top and bottom of the box, respectively. We can see that there is no contribution from the sides of the box, because they are taken to be normal to the surface. In polar coordinates the areas are expressed as and  . Gauss’s law then yields

and now we get the relation

.

This allows us to calculate

Noting that the is the same as E when , this maybe written as

                              (3)

which is analogous to two-dimensional expression.

Going back to the 3-dim problem, we use the same method as above. This time however, the areas of the top and bottom of the Gaussian box are

which in turn yields,

Rearranging gives,

Note that this reduces to the two dimensional expression (3) in cylindrical limit,

Problem 1.5

The time-averaged potential of a neutral hydrogen atom is given by

where q is the magnitude of the electronic charge, and being the Bohr radius. Find the distribution of charge( both continuous and discrete) that will give this potential and interpret your result physically.

Solution:

We can solve the distribution of charge by solving first the charge density using the Poisson’s equation

We are given a hint to find the discrete distribution of charge, meaning likely a delta function will be in our answer.

Now, we solve Poisson’s equation for ,

where .

But . So,

Using the product rule on the first term, we obtain

Then we distribute the term and we get

And then using product rule, we have

After the terms cancel, we are only left with

Using our delta function equation for the first term,

thus

Physically, this is the point charge of the proton nucleus represented by the

delta function at the center of the atom, surrounded by the negative electron

cloud.

Author: Quennie J. Paylaga, Master of Science in Physics student

Problem 1.8 (Chapter 1 of Classical Electrodynamics 3rd Edition by JD Jackson)

Calculate the electrostatic energy (express it in terms of equal and opposite charges Q and -Q placed on the conductors and the potential difference between them) and the energy densities of the electrostatic field of the following capacitors:

(a) two large, flat, conducting sheets of area A, separated by a small distance d;

(b) two concentric conducting spheres with radii a, b (b>a);

(c) two concentric conducting cylinders of length L, large compared to their radii a, b (b>a).

Solution:

Before we can solve for the electrostatic energy and energy densities of the three capacitors, we need to solve first the electric field, potential difference and capacitance of each capacitor. The following formula are the tools in solving this problem.

Electric field E (Gauss Law):

Potential Difference V:

Capacitance C:

Electrostatic Energy W:

Energy Density w:

For Capacitor A

Its electric field is

From Gauss’s Law

The electric field between the two plates of Capacitor A is

where (charge per unit length of the parallel-plate capacitor).

The potential difference of Capacitor A is

The capacitance of Capacitor A is

Thus, the electrostatic energy of  Capacitor A is

and its energy density is

For Capacitor B

The electric field for two conducting spheres is given by:

Its potential difference is

And its capacitance is

Its electrostatic energy is calculated as

and its energy density is given by

For Capacitor C

The electric field for cylindrical capacitor is

Its potential difference is

Its capacitance C is

And its electrostatic energy is given by

and its energy density is

where (charge per unit length of a cylindrical capacitor).

For the parallel-plate capacitor, the energy density is constant and for the spherical capacitor, the energy is more strongly concentrated close to the inner conductor than in the case of a parallel-cylinder capacitor.

Problem 1.4 (Classical Electrodynamics, 3rd Edition by Jackson)

Each of the three charged spheres of radius a has a total charge Q. One is conducting, one has a uniform charge density within its volume and one having a spherically symmetric charge density that varies radially as where (r>-3). Use Gauss’ theorem to obtain the electric fields both inside and outside the sphere.

SOLUTION:

A. Conducting Sphere

A.1 INSIDE

Note that no charge resides inside a conducting sphere. All charges reside in the outer surface thus

A.2 OUTSIDE

Let r be the distance from the center of sphere a.

B. SPHERE WITH UNIFORM CHARGE DENSITY

Since Q is specified to be the total charge then we can get an expression for

B.1 INSIDE SPHERE B

Let b be the distance from the center of the sphere.

b is evaluated from 0 to r where r<a, from 0 to and from o to

B.2 OUTSIDE THE SPHERE

Outside the sphere, the total charge enclosed is still Q.

where r is evaluated from o to a

so that

C. SPHERE WITH

The charge density has the form  where A is a constant. Now the total charge for sphere C should be Q thus

since r is evaluated from o to a

so that

and the charge density can be written as

C.1 INSIDE SPHERE C

C.2 OUTSIDE THE SPHERE

MSPhysics1-MSU-IIT

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Problem 1.10
Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

Proof:

To prove this problem, we are going to use the Green’s  Second Identity which is given by,

Choosing (the scalar potential), and be the integration variable, we have

.       Eq.(1)

Let us solve Eq.(1) term by term. For the first integral,

,   since

,  since if  V contains .

For the second integral,

,  since .

But because there is no charge in the volume that we are integrating(Charge-free) . So the second integral becomes

.

For the third integral,

.

For the fourth integral,

.

But , then

.

Using Divergence Theorem,

,

the fourth integral becomes

.

But and (again, this is true for a charge-free volume! ), then the fourth integral  would be equal to zero, that is,

.

Thus, Eq.(1) is simplified into

.

Hence, the scalar potential is then equal to

.         Eq.(2)

Now, we have proven the mean value theorem.  Eq.(2) says that the potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

——————————————

Use Gauss’s theorem and to prove the following:

(a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.)

Solution:

Suppose that the field were initially nonzero. Since this is a conductor, any charges in the interior would move in response to the field. After a time, this process stops since the moving charges produce currents which dissipate energy. The final configuration would then have charges arranged so that the interior is zero. Recall that in equilibrium, the electric field inside a conductor is zero. Since everywhere inside the conductor, then from Gauss’s Law, , the charge density everywhere in the interior. Therefore, every point inside a conductor has zero charge, and any excess charge can only reside on the surface of the conductor.

(b)  A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it.

Solution:

Part 1. Consider the charge exterior to the conductor which produces an electric field, as shown in the figure.

The electric field in the conductor is zero, with induced charge densities on the exterior and interior surfaces of the conductor.

i) Imagine moving a charge on the interior surface from point A to point B along path 2 which goes throughout the conductor itself. Since in the conductor, along this path.

ii) Move the same charge from A to B along path 1, in the interior cavity of the conductor. Since the electrostatic field is conservative, along its path.

This must be true also for any path in the interior. So generally, in the interior. Therefore the conductor shields its interior from field due to charge placed outside.

Part 2. Consider a positive charge placed inside a hollow conductor as shown in the figure.

The charge induces a charge density in the interior surface of the conductor in such a way that the electric field in the interior of the conductor is zero. Assuming that  the conductor is charge neutral, this means that there is an induced charge density on the exterior surface of total charge Q. If we apply Gauss’s Law to the Gaussian surface G surrounding the conductor, the total charge enclosed is still . Therefore, there is an electric field outside the conductor.

(c) The electric field at the surface of a conductor is normal to the surface and has a magnitude , where is the charge density per unit area on the surface.

Solution:

Note that in equilibrium, the field at exterior surface must be normal to the surface, so that the tangential component is zero. The magnitude of the field is derived using Gauss’s Law with a Gaussian pillbox which cuts through the surface. The electric field is zero on the conducting side of the pillbox. So,

, with the area on the surface.

Rearranging, we get

.

Define as the charge per unit area ,

.

—————————–

Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.

Author: Kayrol Ann B. Vacalares

MS-Physics 1, MSU-Iligan Institute of Technology

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Prove Green’s Reciprocation Theorem:

If is the potential due to a volume-charge density within a volume V and a surface charge density on the  conducting surface S bounding the volume V, while is the potential due to another charge distribution and , then

Solution:

Using Green’s Theorem:

we can replace:

and we can also use the Poisson’s Equation, where we have:

and also the normal derivative of the potential derived from the boundary conditions to yield a surface charge density,

We can use these equations and plug it in Green’s Theorem.

Plugging in:

a.) letting and  

b.) Plugging in Poisson’s Equation, we have:

c.) Plugging in and

cancel out the we get Green’s reciprocation theorem:

To prove this formula, we use the following:

Where: and

Using the equation above:

We can factor out in the first term to give:

Note that for the second term, the permutation of indices are odd, rearranging them to ijk will give the negative:

Thus,

Prove: ___________________________________________________________

Proof: First, we define the following vectors as:

; and

Now, 

if we let i=k, then . Furthermore,

Now, the derivative of orthonormal basis , that is, and the derivative of a coordinate X, . Also, , thus

=

=

=

=

It is noted that . Therefore,

=

This is further equivalent to the ratio of the component of perpendicular to , that is

=

since .

To show their equivalence, we use the BAC-CAB Rule in the definition of . So,

=

=

Thus, we conclude that

_____________________________________________________

where:

Sol’n:

then:

or