Quantum Science Philippines
Quantum Science Philippines

The Normal Derivative Of Electric Field

By Euprime B. Regalado

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

\frac{1}{E}\frac{\partial E}{\partial n}= - \left(\frac{1}{R_1}+\frac{1}{R_2}\right)

where R_1 and R_2 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

0= \int_s\vec{E}\cdot\hat{n}da= E_{top}\nabla_{atop}-E_{bottom}\nabla_{abottom}       (2)

where \nabla_{atop} and \nabla_{abottom} 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 \nabla_{atop}= (R+\epsilon)d\theta dz and \nabla_{abottom}=Rd\theta dz . Gauss’s law then yields

0= E_{top}(R+\epsilon)d\theta dz - E_{bottom}Rd\theta dz

and now we get the relation


This allows us to calculate

\frac{\partial E}{\partial n} = \lim_{\epsilon \to \ 0} \frac{E_{top} -E_{bottom}}{\epsilon} = \lim_{\epsilon \to \ 0} - \frac{E_{top}}{R} = - \frac{E_{top}}{R}

Noting that the E_{top} is the same as E when \epsilon\to\ 0 , this maybe written as

\frac{1}{E}\frac{\partial E}{\partial n}= -\frac{1}{R}                               (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


= (R_1 + \epsilon) (R_2 + \epsilon) d \Omega ,

\nabla_{abottom}= R_1 R_2 d \Omega

which in turn yields,

\frac{\partial E}{\partial n} \\ = \lim_{\epsilon \to \ 0} \frac{E_{top}-E_{bottom}}{\epsilon} \\ = \lim_{\epsilon \to \ 0} - E_{top} (\frac{1}{R_1}+ \frac{1}{R_2}+ \frac{\epsilon}{R_1 R_2}) \\ = - E_{top}(\frac{1}{R_1} + \frac{1}{R_2})

Rearranging gives,

\frac{1}{E} \frac{\partial E}{\partial n} = - (\frac{1}{R_1} + \frac{1}{R_2})

Note that this reduces to the two dimensional expression (3) in cylindrical limit, R_2 \to \infty .



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Solving for the distribution of charge where time-averaged potential is given

by Sim P. Bantayan, MSPhysics I, MSU-IIT


Problem 1.5

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

\Phi = \frac{q}{4\pi\epsilon}\frac{e^{\alpha r}}{r}(1 + \frac{\alpha r}{2})

where q is the magnitude of the electronic charge, and \alpha^{-1} = a_0/2, a_0 being the Bohr radius. Find the distribution of charge( both continuous and discrete) that will give this potential and interpret your result physically.



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

\bigtriangledown^2 \Phi= \frac{\rho}{\epsilon_0}.

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 \rho,

\bigtriangledown^2 \Phi= \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})[\frac{e^{\alpha r}}{r} + \frac{\alpha e^{\alpha r}}{2}]

where \bigtriangledown^2 \equiv \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}).

But \bigtriangledown^2 \Phi= \frac{\rho}{\epsilon_0}. So,

\frac{\rho}{\epsilon_0}= \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})[\frac{e^{\alpha r}}{r} + \frac{\alpha e^{\alpha r}}{2}]

Using the product rule on the first term, we obtain

\rho = -\frac{q}{4\pi}\frac{1}{r^2}\frac{\partial}{\partial r}(e^{-\alpha r}r^2\frac{\partial}{\partial r}(\frac{1}{r}) - \alpha re^{-\alpha r} - \frac{\alpha^2 r^2}{2}e^{-\alpha r})

Then we distribute the \frac{1}{r^2}\frac{\partial}{\partial r} term and we get

\rho = -\frac{q}{4\pi}[\frac{1}{r^2}\frac{\partial}{\partial r}((e^{-\alpha r}r^2\frac{\partial}{\partial r}(\frac{1}{r})) - \frac{1}{r^2}\frac{\partial}{\partial r}(\alpha re^{-\alpha r}) - \frac{1}{r^2}\frac{\partial}{\partial r}(\frac{\alpha^2 r^2}{2}e^{-\alpha r}]

And then using product rule, we have

\rho = -\frac{q}{4\pi}[e^{-\alpha r}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})\frac(1)(r) + \frac{\alpha}{r^2}e^{-\alpha r} + \frac{\alpha^2}{r}e^{-\alpha r} - \frac{\alpha}{r^2}e^{-\alpha r} + \frac{\alpha^3}{2}e^{-\alpha r} - \frac{\alpha^2}{r}e^{-\alpha r}]

After the terms cancel, we are only left with

\rho = -\frac{q}{4\pi}[e^{-\alpha r}\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r})\frac{1}{r} + \frac{\alpha^3}{2}e^{-\alpha r} ]

Using our delta function equation for the first term,

\rho = -\frac{q}{4\pi}[-4\pi\delta(\vec{r}) + \frac{\alpha^3}{2}e^{-\alpha r}],

thus \rho = q\delta(\hat{r}) - \frac{q}{8\pi}\alpha^3e^{-\alpha r}

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






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Electrostatic Energy and Energy Densities of Different Capacitors

Electrostatic Energy and Energy Densities of Different Capacitors

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).


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):

\oint \vec{E} \bullet d \vec{a} = \frac{Q_{enc}}{\epsilon_{o}}

Potential Difference V:

V = \int \vec{E} \bullet d \vec{l}

Capacitance C:

C = Q / V

Electrostatic Energy W:

W = \frac{1}{2} C V^2

Energy Density w:

w = \frac{\epsilon_o}{2} \vert E \vert^2


For Capacitor A

Its electric field is

\int \vec{E} \bullet d \vec{a} = 2 A \vert \vec{E} \vert

From Gauss’s Law

2 A \vert \vec{E} \vert = \frac{Q_{enc}}{\epsilon_{o}} = \frac{\sigma A}{\epsilon_{o}}


\vec{E} = \frac{\sigma A}{2 A \epsilon_{o}} = \frac{\sigma}{2 \epsilon_{o}}

The electric field between the two plates of Capacitor A is

E = \frac{\sigma}{\epsilon_{o}}

where \sigma = \frac{Q}{A} (charge per unit length of the parallel-plate capacitor).

The potential difference of Capacitor A is

V = \int_0^d \vec{E} \bullet d \vec{l} = \int_0^d \frac{\sigma}{\epsilon_{o}} \bullet d \vec{l} = \frac{\sigma d}{\epsilon_{o}} = \frac{Qd}{\epsilon_{o} A}

The capacitance of Capacitor A is

C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{\epsilon_{o} A}} = \frac{A \epsilon_{o}}{d}

Thus, the electrostatic energy of  Capacitor A is

W = \frac{1}{2} C V^2 = \frac{1}{2} \left(\frac{A \epsilon_o}{d}\right) \left(\frac{Qd}{\epsilon_o A}\right)^2 = \frac{1}{2} \frac{Q^2 d}{\epsilon_o A}

and its energy density is

w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{\sigma}{\epsilon_o}\right)^2 = \frac{\sigma^2}{2 \epsilon_o} \hspace{0.5cm} for \hspace{0.3cm} 0 < r < d .


For Capacitor B

The electric field for two conducting spheres is given by:

\vec{E} = \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2} \hat{r}


E = \frac{Q}{4 \pi \epsilon_o r^2}

Its potential difference is

V = - \int_b^a \vec{E} \bullet d \vec{l} = - \frac{Q}{4 \pi \epsilon_o} \int_b^a \frac{1}{r^2} dr


V = \frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)

And its capacitance is

C = \frac{Q}{V} = \frac{Q}{\frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)} = 4 \pi \epsilon_o \frac{ab}{b - a}

Its electrostatic energy is calculated as

W = \frac{1}{2} C V^2 = \frac{1}{2} \left(4 \pi \epsilon_o \frac{ab}{b - a}\right) \left(\frac{Q}{4 \pi \epsilon_o} \left(\frac{1}{a} - \frac{1}{b}\right)\right)^2 = \frac{Q^2}{8 \pi \epsilon_o} \left(\frac{b - a}{ab}\right)

and its energy density is given by

w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}\right)^2 = \frac{Q^2}{32 \pi^2 \epsilon_o} \frac{1}{r^4} \hspace{0.5cm} for \hspace{0.2cm} a<r<b


For Capacitor C

The electric field for cylindrical capacitor is

\vert \vec{E} \vert 2 \pi r L = \frac{Q}{\epsilon_o}


E = \frac{Q}{2 \pi r L \epsilon_o}

Its potential difference is

V = - \int_b^a \vec{E} \bullet d \vec{l} = - \frac{Q}{2 \pi L \epsilon_o} \int_b^a \frac{1}{r} d r = \frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}

Its capacitance C is

C = \frac{Q}{V} = \frac{Q}{\frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}} = \frac{2 \pi L \epsilon_o}{ln \frac{b}{a}}

And its electrostatic energy is given by

W = \frac{1}{2} C V^2 = \frac{1}{2} \left(\frac{2 \pi L \epsilon_o}{ln \frac{b}{a}}\right) \left(\frac{Q}{2 \pi L \epsilon_o} ln \frac{b}{a}\right)^2 = \frac{Q^2}{4 \pi L \epsilon_o} ln \frac{b}{a}

and its energy density is

w = \frac{\epsilon_o}{2} \vert E \vert^2 = \frac{\epsilon_o}{2} \left(\frac{Q}{2 \pi L \epsilon_o} \frac{1}{r}\right)^2 = \frac{Q^2}{8 \pi^2 L^2 \epsilon_o} \frac{1}{r^2} = \frac{\lambda^2}{8 \pi^2 \epsilon_o} \frac{1}{r^2} \hspace{0.5cm} for \hspace{0.2cm} a<r<b

where \lambda = \frac{Q}{L} (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.


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Solving for the electric field using Gauss’ theorem

Bianca Rae B. Sambo

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  r^n where (r>-3). Use Gauss’ theorem to obtain the electric fields both inside and outside the sphere.


A. Conducting Sphere


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

 \oint{E}\bullet{nda}=0 which implies that \vec{E}=0   for r<a



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





\oint{Er^2d\Omega}=\frac{Q}{\epsilon_0} where d\Omega is the solid angle.


{E}{4\pi}r^2 =\frac{Q}{\epsilon_0}


\vec{E}=\frac{Q}{{4\pi}{\epsilon_0}r^2}\hat{r}          where {r}\geq{a}



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

  \rho=\frac{Q}{volume}=\frac{3Q}{{4\pi}a^3}   so that


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

   \oint{E}\bullet{nda}=\frac{1}{\epsilon_0}{\int_v{\rho dV}}=\frac{3Q}{{4\pi}{\epsilon_0}a^3}\int{b^2db}\int{sin\theta{d\theta}}\int{d\phi}

b is evaluated from 0 to r where r<a, \theta from 0 to \pi and \phi from o to 2\pi

   \oint{E{r^2}{d\Omega}}=\frac{3Q}{{4\pi}{\epsilon_0}a^3} {4\pi} \frac{r^3}{3}


   E{4\pi}{r^2}=\frac{3Q}{{4\pi}{\epsilon_0}a^3} {4\pi} \frac{r^3}{3}


  \vec{E} =\frac{Q}{{4\pi}{\epsilon_0}a^3} r\hat{r}     but \vec{r}=r\hat{r}





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

   q_{enc}=\int_v{{\rho}dV}=\frac{3Q}{{4\pi}a^3} {4\pi} \frac{r^3}{3}

where r is evaluated from o to a


so that



  \vec{E}=\frac{Q}{{4\pi}{\epsilon_0}r^2}\hat{r}            where {r}\geq{a}


C. SPHERE WITH {\rho}{\propto}{r^n}

The charge density has the form {\rho}=A{r^n} where A is a constant. Now the total charge for sphere C should be Q thus

   Q=\int{{\rho}dV}=A\int{{r^n}{r^2}{dr}}\int{{sin\theta}d\theta}\int{{d\Omega}}      Q=A\int{{r^{n+2}}{dr}}\int{{sin\theta}d\theta}\int{{d\Omega}}=A\frac{r^{n+3}}{(n+3)}{4\pi}=A{\frac{{4\pi}}{(n+3)}}{a^{n+3}}

since r is evaluated from o to a

so that


and the charge density can be written as

  \rho=\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n   where r is the distance from the center of sphere C



   \oint{E{\bullet}nda}=\frac{1}{\epsilon_0}\int_v{\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n}dV






  \vec{E}=\frac{Q{r^{n+1}}}{{4\pi}{\epsilon_0}{a^{n+3}}}\hat{r}    where r<a



   \oint{E{\bullet}nda}=\frac{1}{\epsilon_0}\int_v{\frac{(n+3)Q}{{4\pi}a^{n+3}} r^n}dV


  \oint{E{cos\theta}da}=\frac{1}{\epsilon_0}{4\pi}\frac{(n+3)Q}{{4\pi}a^{n+3}}\int{r^{n+2}}dr    where r is evaluated from o to a




  \vec{E}=\frac{Q}{{4\pi}{\epsilon_0}{r^2}}\hat{r}   where {r}\geq{a}





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Mean Value Theorem (Classical Electrodynamics)

Roel N. Baybayon



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.



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

\int_V \left(\phi\nabla^2 \psi-\psi \nabla^2 \phi\right)d^3v=\oint_S \left(\phi\frac{\partial\psi}{\partial n}-\psi\frac{\partial \phi}{\partial n}\right)da.

Choosing \phi=\Phi (the scalar potential), \psi=\frac{1}{R} and x' be the integration variable, we have

\int_V \left[\Phi(x')\nabla^2 \left(\frac{1}{R}\right)-\frac{1}{R} \nabla^2  \Phi(x')\right]d^3x'=\oint_S \left[\Phi(x')\frac{\partial}{\partial  n}\left(\frac{1}{R}\right)-\frac{1}{R}\frac{\partial \Phi(x')}{\partial n}\right]d^2x'.       Eq.(1)

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

\int_V \Phi(x')\nabla^2  \left(\frac{1}{R}\right)d^3x'=-4\pi\int_V \Phi(x')\delta(x-x')d^3x',   since \nabla^2 \frac{1}{R}=-4\pi\delta(x-x')

\int_V \Phi(x')\nabla^2  \left(\frac{1}{R}\right)d^3x'=-4\pi\Phi(x),  since \int \delta (x-x')d^3x'=1 if  V contains x=x'.

For the second integral,

-\int_V \frac{1}{R} \nabla^2   \Phi(x') d^3x'=\int_V \frac{1}{R} \frac{\rho}{\epsilon_o} d^3x',  since \nabla^2 \Phi(x')=-\frac{\rho}{\epsilon_o}.

But \rho=0 because there is no charge in the volume that we are integrating(Charge-free) . So the second integral becomes

-\int_V \frac{1}{R} \nabla^2   \Phi(x') d^3x'=0.

For the third integral,

\begin{array}{rcl}\oint_S \Phi(x')\frac{\partial}{\partial   n}\left(\frac{1}{R}\right) d^2x' & = & \oint_S \Phi(x')\left(-\frac{1}{R^2}\right) da\\ & =& -\frac{1}{R^2}\oint_S \Phi(x') d^2x'\end{array}.

For the fourth integral,

-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial  n} d^2x'=-\oint_S\frac{1}{R}\left(\nabla \Phi(x')\cdot \hat{n'}\right) d^2x'.

But \vec{E}=-\nabla \Phi(x'), then

-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial  n}  d^2x'=\oint_S\frac{1}{R}\left(\vec{E}\cdot \hat{n'}\right)  d^2x'.

Using Divergence Theorem,

\int_V (\nabla\cdot\vec{A})d^3x=\oint_S(\vec{A}\cdot \hat{n}) da,

the fourth integral becomes

-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial  n}   d^2x'=\int_V\frac{1}{R}(\nabla\cdot \vec{E})  d^2x'.

But \nabla\cdot\vec{E}=\frac{\rho}{\epsilon_o} and \rho=0 (again, this is true for a charge-free volume! ), then the fourth integral  would be equal to zero, that is,

-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial  n}   d^2x'=0.

Thus, Eq.(1) is simplified into

-4\pi\Phi(x)=-\frac{1}{R^2}\oint_S \Phi(x') d^2x'.

Hence, the scalar potential is then equal to

\Phi(x)=\frac{1}{4\pi R^2}\oint_S \Phi(x') d^2x'.         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.


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