The Normal Derivative Of Electric Field | Quantum Science Philippines
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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

\oint_s\vec{E}\cdot\hat{n}da=0

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

E_{bottom}=E_{top}(1+\frac{\epsilon}{R}).

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

\nabla_{atop}

= (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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