Gauss's Law for Electric Fields

Question 1

The Fundamental Forces

Answer 1

• gravitational forces
• electromagnetic forces
• strong nuclear force
• weak nuclear force

Question 2

Field

Answer 2

• interaction at a distance
• certain objects set up a field around them which other particles interact with
• they are illustrated with arrows and field lines

Question 3

Fields and Flow

Answer 3

-nothing is flowing in a magnetic or electric field, but the mathematical relationships between various quantities in these fields will be the same as in fluid flow

Question 4

Electrostatics and Magnetostatics

Answer 4

• stationary charges have constant electric fields; electrostatics
• steady currents have constant magnetic fields; magnetostatics

Question 5

Steady Currents

Answer 5

• a steady current I, means a continuous flow that has been going on forever without charge piling up anywhere
• a moving point charge cannot constitute a steady current since it is here one instant and gone the next

Question 6

Coulomb’s Law

Electric Field of a Point Charge

Answer 6

|E(|r) = kq/r²r^

k = 1/4πε0 = 8.99x10^9 Nm²/C²

Question 7

Electromagnetism - Superposition Principle

Answer 7

-the net response at a given place is the sum of the responses which would have been caused by each stimulus individually

Question 8

Continuous Charge Distribution

Charge Density

Answer 8

ρ = Q/V
dq = ρdV

Question 9

Continuous Charge Distribution

Electric FIeld

Answer 9

|E(|r) = ∫ kdq/r² r^ = ∫ kρ/r² r^ dV

The electric field is the integral over all charge of kdq/r² or the integral over the entire volume of kρ/r².

Question 10

Gauss’s Law for Electric Fields - Integral Form

Answer 10

∮(|E . n^)dA = Q/ε0

The integral of electric field in the direction perpendicular to the surface across a closed surface is equal to the total charge enclosed by that surface divided by the permitivity of free space
OR
An electric charge produced an electric field and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface.

Question 11

Solid Angle

Answer 11

ΔA’ = r² x ΔΩ

where A’ is the are on the surface of the sphere
r is the radius of the sphere and Ω is the solid angle

• to go right around a sphere or any closed surface you have to travel around by Ω=4π steradians
• this is why the surface area of a sphere is 4πr²

Question 12

Gauss’s Law for Electric Fields - Integral Form

Multiple Point Charges

Answer 12

∮(|E . n^)dA = ∮ Σ|E.n^ dA = ∮|E1.n^dA + ∮|E2.n^dA + …
=q1/ε0 + q2/ε0 + …
=Q/ε0

Question 13

Gauss’s Law for Electric Fields - Integral Form

Charge Outside of the Closed Surface

Answer 13

-the net flux from any charge outside of the closed surface is zero as the flux into the surface from that charge is equal to the flux out, so net flux is zero.

Question 14

Gauss’s Law for Electric Fields

Sources and Sinks

Answer 14

-net flux will tell you whether inside a closed surface you have a net source or sink of electric field lines and whether the net flow is out or in

Question 15

Is a positive charge a source or a sink?

Answer 15

-positive charge acts as a source of field lines like a tap within the enclosed volume generating the flow out of the surface containing that volume

Question 16

Is a negative charge a source or a sink?

Answer 16

-negative charge acts as a drain of field lines, like a sink within the enclosed volume generating a flow into the surface containing that volume

Question 17

Electric Fields - Definition of Divergence

Answer 17

div(|E) = ∇ . |E =
V->0 lim 1/ΔV ∮ |E . n^dA

-this limit gives the next flux per unit volume at a point
|∇ . |E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z

Question 18

Gauss’s Law for Electric Fields

Differential Form

Answer 18

|∇ . |E = ρ/ε0

Question 19

Gauss’s Law for Electric Fields

Integral to Differential Form

Answer 19

-start with Gauss's Law:
∮(|E . n^)dA = Q/ε0
-divide both sides by ΔV
1/ΔV*∮(|E . n^)dA = Q/ΔV *1/ε0
-the left hand side is equal to ∇ . |E
-right hand side is equal to ρ/ε0, where ρ is the charge density
-this gives the differential form of Gauss's Law
∇ . |E = ρ/ε0

Question 20

Gauss’s Law for Electric Fields

Description of Differential Form

Answer 20

• the electric field produced by electric charge diverges from positive charge and converges upon negative charge
• the only places at which divergence of the electric field is not zero are the locations at which charge is present

Question 21

Where does divergence or convergence occur in an electric field?

Answer 21

• the only places where divergence or convergence occurs is at point charges
• even if at a point in the field the lines appear to be moving away from each other, if you took a small volume the next flux in and out would still be 0 so by definition, no divergence or convergence

Question 22

Electric Flux Through a Closed Surface S

|E uniform and perpendicular to S

Answer 22

φ = |E| * (surface area)

Question 23

Electric Flux Through a Closed Surface S

|E uniform and at an angle to S

Answer 23

φ = |E . ^n

Question 24

Electric Flux Through a Closed Surface S

|E non-uniform and at variable angles to S

Answer 24

φ = ∫ |E . ^n dA

Question 25

Rule for Drawing Electric Fields

Answer 25

1) electric field lines must originate on positive charge and terminate on negative charge
2) the net electric field at any point is the vector sum of all electric fields present at that point
3) electric field lines can never cross since that would indicate that the field point in two directions at the same location which violates rule 2
4) electric field lines are always perpendicular to the surface of a conductor in equilibrium

Question 26

Electric Field of a Conducting Sphere

Answer 26

|E = 1/4πεo * Q/r² * ^r
-outside the sphere a distance r from the centre

|E = 0
-inside

Question 27

Electric Field of a Uniformly Charged Insulating Sphere (radius=ro)

Answer 27

|E = 1/4πεo * Q/r² * ^r
-outside the sphere, a distance r from the centre

|E = 1/4πεo * Qr/(ro)³ * ^r
-inside the sphere, a distance r from the centre

Question 28

Electric Field of an Infinite Line Charge

linear charge density=λ

Answer 28

|E = 1/2πεo * λ/r * ^r

-a distance r from the line

Question 29

Electric Field of an Infinite Flat Plane

surface charge density = σ

Answer 29

|E = σ/2εo ^n

Question 30

Uses of the Integral Form of Gauss’s Law for Electric Fields

Answer 30

1) given information about a distribution of electric charge, you can find the electric flux through a surface enclosing that charge
2) given information about the electric flux through an enclosed surface you can find the total electric charge enclosed by that surface

Question 31

Insulator/Dielectric

Definition

Answer 31

-in a dielectric material, charges do not move freely but can be displaced from their equilibrium position

Question 32

Special Gaussian Surface

Answer 32

• in certain cases a surface can be constructed such that it is possible to calculate the electric field
• such a surface must meet two conditions:
  1) the electric field must be either parallel or perpendicular to the surface normal at all points on the surface
  2) the electric field must be constant or zero over the surface

Question 33

Describe the fundamental difference between the integral form and the differential form of Gauss’s law for electric fields

Answer 33

-the differential form deals with divergence of the electric field and charge density at INDIVIDUAL POINTS in space whereas the integral form entails the integral of the normal component of the electric field OVER A SURFACE

Question 34

Uses of the Differential Form of Gauss’s Law for Electric Fields

Answer 34

1) if spatial variation of the vector electric field is known at a specified location, then you can find the charge density at that location
2) if volume charge density is known, the divergence of the electric field may be determined