Student Reader

Electricity and Magnetism

Term Units Overview
Force (F)

Force from a charge: F = E q

Two charges q1 and q2 exert a force on each other: F12 = k q1 q2 / r122
Electric Field (E) Volts / meter

E vector = E0 ( x , t ) yˆ → E is y direction → E0 sin ( k x - ω t ) y

Electric field from a charge: E = F / q

Electric field of a point charge: E = k q / r2 (where r is distance from charge)
Peak Electric Field (E0) Volts / meter E0 = B0 c = µ0 ε0 ω c
Magnetic Field (B) Tesla B vector = B0 ( x , t ) zˆ → B is z direction → B0 sin (k x - ω t ) z B = B0 cos ( k x - ω t ) B = µ0 ε0 ω cos ( k x - ω t )
Peak Magnetic Field (B0) Tesla (T) B0 = c / E0 B0 = µ0 ε0 ω
Charge (q or Q) Coulumbs (C)
Electric Potential (V)

Potential energy per charge.

Electric potential by some source point charge: V = k q / r

Electric potential between two point charges: V = k q1 / r1 + k q2 / r2
Light Speed (c) 3 × 108 m / s

c = E0 / B0 = k / ω

c = v = ƒ λ

c2 = 1 / ( µ0 ε0 )
Energy Density (u) J m-3

uE = ½ ε0 E2

uB = ½ B2 / µ0
Poynting Vector (S) J m-2 s-1 Rate at which wave carries energy across unit area per unit time. Average S = ½ ε0 c E02 = ½ E0 B0 / µ0 = Erms Brms / µ0 = Erms2 / ( µ0 c ) = Brms c / µ0
Radiation Pressure (P) Pressure equals the average Poynting vector times α and divided by c. It is independent of area.

P = α S / c

α is the coefficient of reflection and refers to a materials ability to absorb or reflect. 1 (fully absorbing) < α < 2 (fully reflecting). A common material in homework problems is a metal plate, which is assumed to fully reflect and thus have α = 2.
Power Power = Intensity × Area
Power Received

Consider a spherical lightbulb with power output ℘; what is the power received by another sphere with radius R at distance D from the lightbulb? (A common homework problem involves the sun and a planet.) The power received is ID π r2. This is the intensity of the light at distance D ( ℘ / ( 4 π D2 ) times the area of a circle with radius R ( π R2 ). The object receiving the power output is treated as a circle, because the shadow it casts is a simple two-dimensional circle. Otherwise you would have to treat it as a sphere, and then integrate based on the power received at various angles along the sphere's surface.

For a sphere with radius R at a distance D from the light source: ℘received = ( ℘ R2 ) / ( 4 D2 )

Consider a spherical light bulb with power output ℘. Its intensity is its power output ℘ divided by its surface area (4πr2). At 1 meter away, its intensity is ℘ divided by the surface area of a sphere with radius 1 meter. Note that the power has not gone down at 1 meter, 2 meter or however many meters away. However, its intensity has gone down at increasing distances; the power is spread over a greater surface area.

I = ℘ / area


Term Description
Energy Constant = k 8.99 x 109 N m2 C-2
Permittivity Constant ε0 8.8542 x 10-12 C2 N m-2
Magnetic Constant = µ0 4 π × 10−7 V s / ( A m )

Directions for Electrogmagnetic Waves

k is parallel to v, the propagation direction
k is perpendicular to E and to B
E is perpendicular to B
Form the letter L with your right hand. Stick your middle finger at yourself. Your index finger (pointing up) is E. Your thumb (pointing right) is k. Your middle finger (pointing at you) is B.

Doppler Effect for Electromagnetic Waves

ƒ = ƒ0 [ ( c - v ) / ( c + v ) ]½
ƒ0 is the emitted frequency; ƒ is the shifted frequency

Electricity and MagnetismComments