Question

Gauss' Law and Equipotentials A two-di mensional representation of co-centric spheres and coaxial cables looks the same as below (o). Is this 2-D drawing (a) below for a set of spheres (b) or cyl (a What is your initial guess/prediction? You will measure the electrostatic potential for the 2-D drawing and then compare the data to the predicted potential for the sphere and cylinder configurations above. There are two sections to the lab: calculating/predicting the electrie fields in the two configurations (and thus the electrostatic potentials) and making the measurements Calculation: Use Gauss' law to calculate the electric field in between co-centric spheres with inner radius a and outer radius b. Assume the inner sphere has a charge of+Q and the outer shell has a charge of -Q (so the configuration is electrically neutral). Gauss' Law: E. dA = 0% Gaussian surface: Draw a spherical (mimic the symmetry) Gaussian surface inside the outer sphere and label the radius r What is the charge enclosed in the Gaussian surface? What is the direction of Eon the Gaussian surface? What is the direction of d.A on the Gaussian surface? What is E·dA? what is the cosine of the angle between E and dA?

Answer 1

For the case of the internal sphere (ra)

Choose a spherical Gaussian surface of radius r, concentric with the sphere,
as shown in figure 1.

Gauss's law is

$\Phi =\oint \vec{E}.d\vec{A}=\frac{qin}{\epsilon o }$

By symmetry, E is constant everywhere on the surface and leaves the integral

$E\oint dA=E(4\pi r^{2})=\frac{Q}{\epsilon o }$

when solving for E

$E=\frac{Q}{4\pi \epsilon o r^{2}}$

This field is identical to that of a point load.

Figure 2 shows that the electric field lines are normal to the Gaussian surface.

Gaussian surface

the direction of $d\vec{A}$ is also normal to the Gaussian surface and parallel to $\vec{E}$.

$d\vec{A}$ and $\vec{E}$ represent a point product and is given by:

$\vec{E}.\vec{dA}=EdAcos\Theta$

since $\vec{E}$ and $d\vec{A}$ are parallel they form an angle $\Theta=0$, then

$\vec{E}.\vec{dA}=EdAcos(0)=EdA$