# Wright State University Lake Campus/2017-9/Phy2410/Help with Gauss' law

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Note: GS=Gaussian Surface and EC=enclosed charge. R = charged object r=field point.

The quiz at Special:permalink/1391093 is about a cylinder or a sphere. There are actually two cylinders, one a long wire, and the other a flat plane not unlike that of a parallel plate capacitor. It relates a surface integral to a volume integral:

$\varepsilon _{0}\underbrace {\int \int } _{area}{\vec {E}}\cdot d{\vec {A}}=\underbrace {\int \int \int } _{volume}\rho \;dV=Q_{enclosed}$ 1. The first is a surface integral of ${\vec {E}}\cdot {\vec {n}}$ over a closed and the second is a volume integral of the charge density over the volume enclosed by this "closed surface". A "closed surface" is any surface without a "hole": A balloon is an open surface unless you tie off the exit, in which case it is closed.
2. The "outward" unit normal ${\hat {n}}$ can only be defined for a closed surface. We shall call our closed surface a "Gaussian Surface" (GS). Gauss' Law holds for any GS that you can imagine. But it is only useful if the integral is easy to solve.
3. To use Gauss' Law, arrange for the surface to coincide with the "field point", which is defined as the point where you want to calculate the field. We shall only use this law in cases where the integrand is uniform over either all or part of the area. Also, we restrict ourselves to cases where the charge density $\rho$ is uniform. With these simplifications, the integral form of Gauss' law becomes an algebraic expression:  $\varepsilon _{0}EA^{*}=V^{*}\rho$ .
4. We define small-r $r$ to represent the location of the field point, and big-R $R$ to represent the object. This works in spherical coordinates and in cylindrical coordinates if the cylinder is a long wire of radius $R$ . For this wire, $r if the field point is inside the wire, and $r>R$ if the field point is outside the wire.
5. In cylindrical coordinates for a "pancake", $R\rightarrow \infty$ , and the field point is $z$ for a wire of length $H>>R$ . The field point is inside the object if $z and outside the object if $z>H/2$ .

Sphere:   $V_{CE}^{*}={\tfrac {4}{3}}\pi r^{*3}\qquad A^{*}=4\pi r_{GS}^{*2}\quad ({\text{note that }}A=dV/dr)$ • If the field point is inside the object, then both r values are the same and equal to the field point: $r^{*}=r$ • If the field point is outside the object, then $r^{*}=r$ for the surface integral (calculation of $A^{*}$ ) but the smaller value of $r^{*}=R$ for the calculation of the volume integral $A^{*}$ .

Long wire: H>>R:   $V_{CE}^{*}=H\;\pi r^{*2}\qquad A^{*}=H\;2\pi r^{*}$ (V is height × footprint; A is height × circumference.)

(Note that the circumference of a circle is the derivative of its area. Note also that E does not depend on the length of the wire $H$ because that term cancels on both sides.)

• If the field point is inside the object, then both $r^{*}$ values are the same and equal to the field point: $r^{*}=r$ • If the field point is outside the object, then $r^{*}=r$ for the surface integral (calculation of $A^{*}$ ) but the smaller value of $r^{*}=R$ for the calculation of the volume integral $A^{*}$ .

Flat pancake:H<<R:   This problem is tricky, because the pancake has two areas (circles). The electric field is parallel to the unit normal because it points away from positive charge. Hence the area used is twice the area of a circle. If the GS is situated inside the pancake, the volume also doubles, because our GS covers a volume extending from $-z$ to $+z$ (in other words, the "cylinder" is 2z long if z<H/2, but it is H long if z>H/2). We include a dagger on the two $(2^{\dagger })$ as a reminder that we need to double certain numbers since the GS is a cylinder situated at the origin (extending from −z to +z) with two circular areas at each end.

$V_{CE}^{*}=H\;\pi R^{*2}$ (outside object) .. or .. $V_{CE}^{*}=\left(2^{\dagger }z\right)\;\pi R^{*2}$ (inside object)     $A_{GS}^{*}=2^{\dagger }\pi R^{2}$ 