Quantum Science Philippines
Quantum Science Philippines

Proving properties of electric fields using Gauss’s Theorem


Use Gauss’s theorem and \oint \vec{E} \cdot d\vec{l} = 0 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.)


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 \vec{E} = 0 everywhere inside the conductor, then from Gauss’s Law, \oint \vec{E} \cdot \hat{n} da =4\pi \int_V \rho(\vec{x}) d^3 x = o, the charge density \rho (\vec{x}) = o 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.


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 \vec{E} =0 in the conductor, \int_2 \vec{E} \cdot d\vec{l} = 0 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, \vec{E} \cdot d\vec{l} = 0 along its path.

This must be true also for any path in the interior. So generally, \vec{E} =0 in the interior. Therefore the conductor shields its interior from field due to charge placed outside.

Part 2. Consider a positive charge Q 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 Q. 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 \sigma / \epsilon_0, where \sigma is the charge density per unit area on the surface.


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,

\oint \vec{E} \cdot \hat{n} da =EA , with A the area on the surface.

E A = (\frac{q}{\epsilon_0} )

Rearranging, we get

E =(\frac{q}{A} ) (\frac{1}{\epsilon_0} ).

Define \sigma as the charge per unit area q/A,

E = (\frac{\sigma}{\epsilon_0} ).


Classical Electrodynamics, John David Jackson, 3rd Edition, Chapter 1.
Introduction to Classical Electrodynamics, David Griffiths, Chapter 2.
University Physics, Young and Freedman, 11th Edition, Chapter 24.
Faraday’s cage, wikipedia.com.
Gauss’s Law, wikipedia.com.


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

Posted in Electrodynamics, Quantum Science Philippines | 4 Comments »

Prove Green’s Reciprocation Theorem

Author: Kayrol Ann B. Vacalares

MS-Physics 1, MSU-Iligan Institute of Technology



Prove Green’s Reciprocation Theorem:

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

\int_v \rho \Phi' d^{3}x + \int_s \sigma \Phi' da = \int_v \rho' \Phi d^{3}x + \int_s \sigma' \Phi da



Using Green’s Theorem:

\int_v (\phi \nabla^{2} \psi - \psi \nabla^{2} \phi) d^{3}x = \oint_s [\phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi}{\partial n}] da

we can replace:

\phi to \Phi  and \psi to \Phi'

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

\nabla^{2} \Phi = - \frac{\rho}{\epsilon_0} and

\nabla^{2} \Phi' = - \frac{\rho'}{\epsilon_0}

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

\sigma = \epsilon_0 \frac{\partial \Phi}{\partial n} \sigma' = \epsilon_0 \frac{\partial \Phi'}{\partial n}

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

Plugging in:

a.) letting \psi = \Phi' and  \phi = \Phi

\int_v (\Phi \nabla^{2} \Phi') - \Phi' \nabla^{2} \Phi) d^{3}x = \oint_s [\Phi \frac{\partial \Phi'}{\partial n} - \Phi' \frac{\partial \Phi}{\partial n}] da \int_v (\Phi' \nabla^{2} \Phi d^{3}x + \oint_s \Phi' \frac{\partial \Phi}{\partial n} da = -\int_v \Phi \nabla^{2} \Phi' d^{3}x + \oint_s \Phi \frac{\partial \Phi'}{\partial n} da


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

\int_v \Phi (\frac{\rho}{\epsilon_0}) d^{3}x + \oint_s \Phi' \frac{\partial \Phi}{\partial n} da = \int_v \Phi (\frac{\rho'}{\epsilon_0} d^{3}x + \oint_s \Phi (\frac{\partial \Phi'}{\partial n} da


c.) Plugging in \sigma and \sigma'

\int_v \Phi' \frac{\rho}{\epsilon_0} d^{3}x + \oint_s \Phi' \frac{\sigma}{\epsilon_0} da = \int_v \Phi \frac{\rho'}{\epsilon_0} d^{3}x + \oint_s \Phi \frac{\sigma'}{\epsilon_0} da

cancel out the \epsilon_0 we get Green’s reciprocation theorem:


\int_v \rho \Phi' d^{3}x + \int_s \sigma \Phi' da = \int_v \rho' \Phi d^{3}x + \int_s \sigma' \Phi da


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Curl of the product of a scalar and a vector using Levi-Civita

\vec{\nabla} \times f \vec{A}=f \vec{\nabla} \times \vec{A} \ + \ \vec{\nabla} f \times \vec{A}

By Eliezer Estrecho

To prove this formula, we use the following:

\vec{\nabla} \times \vec{A}=\epsilon _{ijk} \hat{e}_{i} \nabla_{j} A_{k}

Where: \vec{\nabla}=\nabla_{j} \hat{e}_{j} and \vec{A}=A_{k} \hat{e}_{k}

Using the equation above:

  \vec{\nabla} \times f \vec{A} =\epsilon _{ijk} \hat{e}_{i} \nabla_{j} f A_{k} \\  \vec{\nabla} \times f \vec{A} =\epsilon _{ijk} \hat{e}_{i} \ (f \nabla_{j} A_{k} + A_{k}\nabla_{j}f) \ \ \ \ \ \ \ product \ rule\\  \vec{\nabla} \times f \vec{A} =\epsilon_{ijk}\hat{e}_{i}f \nabla_{j} A_{k} + \epsilon_{ijk}\hat{e}_{i} A_{k} \nabla_{j} f

We can factor out  f in the first term to give:

f \epsilon_{ijk}\hat{e}_{i} \nabla_{j} A_{k}=f \vec{\nabla} \times \vec{A}

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

\epsilon_{ijk}\hat{e}_{i} A_{k} \nabla_{j} f = -\epsilon_{ijk}\hat{e}_{i} A_{j} \nabla_{k} f = -\vec{A} \times \vec{\nabla}f = \vec{\nabla}f \times \vec{A}


\vec{\nabla} \times f \vec{A}=f \vec{\nabla} \times \vec{A} \ + \ \vec{\nabla} f \times \vec{A}

About the author: Eliezer Estrecho is currently a MS Physics student of MSU-IIT.

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Proving Vector Identity Involving the Unit Vector Using the Levi-Civita and the Kronecker Delta

*author: Michelle R. Fudot


Prove: (\vec{a}.\vec{\nabla)}\hat{n}) =\frac{1}{r}[\vec{a}-\hat{n}(\vec{a}.\hat{n}]=\frac{a_\perp}{r} ___________________________________________________________

Proof: First, we define the following vectors as:

\vec{a}=a_i \widehat{e_i};

\hat{n}=\frac{\vec{X}}{|X|}=\frac{X_j\widehat{e_j}}{|X|}; and

\vec{\nabla} = \partial_k\widehat{e_k}

Now, (\vec{a}.\vec{\nabla})\hat{n}=(a_i \widehat{e_i}\bullet\partial_k\widehat{e_k})\frac{X_j\widehat{e_j}}{|X|}


if we let i=k, then \delta_{ik} = \delta_{ii} = 1. Furthermore,

= a_i\partial_i\frac{X_j\widehat{e_j}}{|X|} = a_i\lgroup\frac{1}{|X|}\partial_iX_j\widehat{e_j} + X_j\widehat{e_j}\partial_i(\frac{1}{|X|})\rgroup = a_i\lgroup\frac{1}{|X|}\lgroup\widehat{e_j}\partial_iX_j+X_j\partial_i\widehat{e_j}\rgroup+X_j\widehat{e_j}\partial_i(\frac{1}{|X|})\rgroup

Now, the derivative of orthonormal basis \widehat{e_j}, that is, \partial_i\widehat{e_j}=0 and the derivative of a coordinate X, \partial_iX_j = \delta_{ij}. Also, \partial_i(\frac{1}{|X|})=-\frac{X_j\widehat{e_j}}{|X|^3}, thus





It is noted that |X| = r. Therefore,


This is further equivalent to the ratio of the component of a perpendicular to \vec{X}, that is

= \frac{a_\perp}{r}

since a_\perp = \hat{n}\times(\hat{n}\times\vec{a}).

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

\frac{a_\perp}{r} = \frac{\hat{n}(-\hat{n}.\vec{a}-\vec{a}(-\hat{n}.\hat{n})}{r}



Thus, we conclude that

(\vec{a}.\vec{\nabla})\hat{n} =\frac{1}{r}(\vec{a}-\hat{n}(\vec{a}.\hat{n})) =\frac{a_\perp}{r}


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Vector Analysis














\vec{a}\cdot(\vec{b}\times\vec{c})=a_i\widehat{e_i}\cdot b_jc_k\in_{jkl}\widehat{e_l}










=b_j\widehat{e_j}\cdot c_ka_i\in_{kij}\widehat{e_j}




=c_k a_ib_j\in_{ijk}


=c_k\widehat{e_k}\cdot a_ib_j\in_{ijk}\widehat{e_k}




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