prelecture_04_guass_law
- [ \oint\limits_{surface}{ \vec E \cdot d \vec A} = \frac{ q_{enclosed}}{ \epsilon_0} ]
- First discuss solid insulator followed by
- Solid Conductor
- Cylindrical Conductor
- Infinite Sheet of Charge
- Casses with Symmetry [ E] may be pulled out of Gauss' Law such that
- [E \oint\limits_{surface}{ d A} = \frac{ q_{enclosed}}{ \epsilon_0}]
Consider a sphere, an infinitely long cylinder, and a plane of infinite length and width (a, b and c below). Imagine that you can hover above each one in your own personal helicopter. In which case do you have the most freedom to move about without your view of the object changing? In other words, for each case consider if there are directions that you can move in without the objects distance or orientation, relative to you, changing.
- The infinite plane offers the most freedom to move without your view changing.
- You can move in two dimensions - any direction parallel to the plane - as you fly above it, and it will still look exactly the same from your perspective. By the same kind of reasoning you will see that you can move in just one dimension above the cylinder - parallel to its length - if you want it to look the same, and you can not move at all above the sphere without your view changing.
- Insulator \Rightarrow charges cannot move
- Let
- [Q ] be the charge
- [a ] be the radius
- [ E = \frac{ Q}{ 4 \pi \epsilon_0 r^2}]
A solid sphere of radius [R] has a total charge [ Q] distributed evenly throughout its volume. What fraction of the total charge is located inside a radius [ \frac{ R}{ 2} ]?
- [ \frac{ 1}{ 8} ]
- Since the charge is uniformly distributed throughout the volume of the sphere, the fraction of the charge that resides inside a given radius is simply equal to the fraction of the spheres volume inside that radius. Since the volume of a sphere is [ 4 \pi \frac{ R^3}{ 3} ], reducing the radius by [\frac{ 1}{ 2}] will reduce the volume by [ \left(\frac{ 1}{ 2}\right)^3 = \frac{ 1}{ 8}].
- Insulator \Rightarrow charges cannot move
- Let
- [Q ] be the charge
- [a ] be the radius
- [r ] distance of the test point
- Charge Density
- [ \rho = \frac{ Q}{ \frac{ 4}{ 3} \pi a^3} ]
- [r > a]
- [ E = \frac{ Q}{ 4 \pi \epsilon_0 r^2}]
- [r < a ]
- [ E = \frac{ \rho}{ 3 \epsilon_0} r ]
Consider a sphere of radius [ R], surface area [ A] and volume [ V]. Suppose you double the radius to [2R]. How does the new surface area [A_{new}] and the new volume [V_{new}] compare to the old values?
- [A_{new} = 4A, V_{new} = 8V]
- The area of a sphere is given by [ A = 4 \pi R^2] and the volume of a sphere is given by [ V = \frac{ 4 \pi R^3}{ 3}]. Doubling [ R] in each case will cause the volume to increase by a factor of [ 2^3 = 8] and the area to increase by a factor of [ 2^2 = 4].
- Solid Conductor
- Charge is on the surface
- Charge free to move
- Inside the sphere
- [ E = 0]
-
Let
- [ Q_{inner} = -q_0 ]
-
Then
- [ \sigma_i = \frac{ -q_0}{ 4 \pi R_i^2 ]
- Induced Inner Charge Density
- [ \sigma_i = \frac{ -q_0}{ 4 \pi R_i^2 ]
-
Let
- [ Q_{Outer} = Q + q_0 ]
-
Then
- [ \sigma_0 = \frac{ Q + q_0}{ 4 \pi R_0^2}]
- Outer Charge Density
- [ \sigma_0 = \frac{ Q + q_0}{ 4 \pi R_0^2}]
- [ E = \frac{ \lambda}{ 2 \pi \epsilon_0 r}
- [ \lambda \Rightarrow] linear charge density
- [ E = \frac{ \rho}{ 2 \epsilon_0 } ]
You are told to use Gauss' law to calculate the electric field near an infinite sheet. The catch is that you have to choose a Gaussian surface different from the one used in the previous example. Which of the following Gaussian surfaces is best suited for this purpose?
- a sphere centered on the plane
- a cylinder with its axis parallel to the plane
- a box with two sides parallel to the plane
- This one!
- The sides of the box are either perpendicular or parallel to the field, so this Gaussian surface is just as good as the perpendicular cylinder used in the above example.
-
[ \oint\limits_{surface}{ \vec E \cdot d \vec A} = \frac{ q_{enclosed}}{ \epsilon_0} ]
-
Let
- [ Q_{inner} = -q_0 ]
-
Then
- [ \sigma_i = \frac{ -q_0}{ 4 \pi R_i^2 ]
- Induced Inner Charge Density
- [ \sigma_i = \frac{ -q_0}{ 4 \pi R_i^2 ]
-
Let
- [ Q_{Outer} = Q + q_0 ]
-
Then
- [ \sigma_0 = \frac{ Q + q_0}{ 4 \pi R_0^2}]
- Outer Charge Density
- [ \sigma_0 = \frac{ Q + q_0}{ 4 \pi R_0^2}]