Some Measures of Electric Charge Name Symbol SI

Some Measures of Electric Charge Name Symbol SI Unit Charge q C Linear Charge Density C/m Surface Charge Density C/m 2 Volume Charge Density C/m 3

Electric Field Due to a Ring of Charge The ring of uniform charge has a linear charge density of (in coulomb/m). Find the electric field strength at point P which is located a distance x from the center of the ring as shown. You must consider the contribution each differential piece (d. Q) has on the electric field at point P. d. Q = ds

Solving for E The vertical component of E will cancel out, while the vertical component has a magnitude of d. E cos.

The Integral

Electric Field Due to a Charged Disk The disk of uniform charge has a surface charge density of (in C/m 2). Find the electric field strength at point P which is located a distance z from the center of the disk as shown. P z The plan is to divide the disk into concentric flat rings, and to add up the contributions each ring has to the electric field at point P. dr r R

Solving for E P z Electric field due to a ring of charge. r dr R

Evaluating the Integral P Let U = (z 2 +r 2) and d. U = 2 r dr. z r R

Infinite Sheet of Charge z + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + The electric field strength at any distance z from an infinite sheet of charge is constant.

Calculate the Electric Field Strength at Point P

Gauss’ Law What is it and how can it help me?

What is Gauss’ Law? ? ? Electric Flux Through a Closed Surface Charge Enclosed Gauss’s Law, like Coulomb’s Law can help you to evaluate the electric field by relating the electric field at a location to the charge(s) creating it. (eo = the permittivity of free space 8. 854 x 10 -12 C 2/(N m 2) = 1/4 k)

Electric Flux The Electric Flux FE is the product of component of the electric field strength E that is perpendicular to a surface times the surface area. (SI: V. m = N. m 2/C) Unit vector perpendicular to A

The Surface Integral Gaussian Surface

Visualizing Electric Flux

What Makes Electric Flux? ? You can think of electric flux as the number of electric field lines that pass through a surface. You can imagine that the more dense the electric field lines, the greater the electric field flux.

More Importantly… The net flux through a closed surface is proportional to the charge enclosed by that surface.

AND, Therefore… If there is no net charge inside some volume of space then the electric flux over the surface of that volume is always equal to zero. If the electric field is parallel to the Gaussian surface, the electric flux is through that surface is zero.

Using Gauss’ Law can be used to evaluate the electric field strength in an electric field, by taking advantage of the symmetry of a charge distribution. The skill in using Gauss’ Law comes in picking a Gaussian surface such that E everywhere on that surface is constant, so that you can pull E out of the integral, and evaluate (1) the area of the Gaussian surface and (2) the enclosed charge (qencl).

A Derivation of Coulomb’s Law For a point charge q, the electric field strength E at distance r from the charge may be found using Gauss’s Law: r + By using a spherical Gaussian surface, we recognize that the electric field strength E at distance r will be constant, and therefore, E falls out of the integral. That’s what we want to happen. The surface integral for a spherical Gaussian surface is the surface area of a sphere.

Recipe for Success 1. Carefully draw a figure showing the location of all charges, the direction of the resulting electric field vectors E and the given dimensions. 2. Draw an imaginary closed Gaussian surface so that the value of the magnitude of the electric field, E, is constant on the surface… you’ll be dealing with spheres, cylinders or other shapes that have some symmetry. 3. Write Gauss’ Law and perform the surface integral, which should be a friendly integral if you choose the right Gaussian surface. 4. You can factor the electric field out of the integral, and simply evaluate the surface integral. 5. Write an expression for qencl, and then solve the Gauss’ Law for the magnitude of the electric field.