Flashcards in Chapter 23 Deck (15):

1

## 23-1

### The electric flux oIo through a surface is the amount of electric field that pierces the surface.

2

## 23-1

### The area vector dA_vector for an area element (patch element) on a surface is a vector that is perpendicular to the element and has a magnitude equal to the area dA of the element.

3

## 23-1

###
The electric flux doIo through a patch element with area vector dA_vector is given by a dot product:

doIo = E_vector (dot) dA_vector .

4

## 23-1

###
The total flux through a surface is given by oIo = S_integral

E_vector (dot) dA_vector (total flux) where the integration is carried out over the surface.

5

## 23-1

###
The net flux through a closed surface (which is used in Gauss’ law) is given by oIo = So_circle_in_integral

E_vector (dot) dA_vector (net flux) where the integration is carried out over the entire surface.

6

## 23-2

###
Gauss’ law relates the net flux oIo penetrating a closed surface to the net charge q_enc enclosed by the surface:

E0*oIo = q_enc (Gauss’ law).

7

## 23-2

###
Gauss’ law can also be written in terms of the electric field piercing the enclosing Gaussian surface: E0*So_circle_in_integral

E_vector (dot) dA_vector (net flux) = q_enc (Gauss’ law).

8

## 23-3

### An excess charge on an isolated conductor is located entirely on the outer surface of the conductor.

9

## 23-3

### The internal electric field of a charged, isolated conductor is zero, and the external field (at nearby points) is perpendicular to the surface and has a magnitude that depends on the surface charge density o-: E = o- / E0

10

## 23-4

###
The electric field at a point near an infinite line of charge (or charged rod) with uniform linear charge density lambda is perpendicular to the line and has magnitude

E = lambda / 2*pi*E0*r (line of charge),

where r is the perpendicular distance from the line to the point.

11

## 23-5

###
The electric field due to an infinite nonconducting sheet with uniform surface charge density o- is perpendicular to the plane of the sheet and has magnitude

E = o- / 2*E0 (nonconducting sheet of charge).

12

## 23-5

###
he external electric field just outside the surface of an isolated charged conductor with surface charge density o- is perpendicular to the surface and has magnitude

E = o- / E0 (external, charged conductor).

Inside the conductor, the electric field is zero.

13

## 23-6

###
Outside a spherical shell of uniform charge q, the electric field due to the shell is radial (inward or outward, depending on the sign of the charge) and has the magnitude

E = (1 / 4*pi*E0) * (q / r^2) (outside spherical shell),

where r is the distance to the point of measurement from the center of the shell. The field is the same as though all of the charge is concentrated as a particle at the center of the shell.

14

## 23-6

### Inside the shell, the field due to the shell is zero.

15