• Contact
• About

  Search by keyword:  

Last modified: January 26, 2018

Electric Fields
---------------
Electric field is defined as the electric force per unit charge. Thus:

E = F/q or F = qE

Point charge:

 E = kq/r2

The direction of the field is taken to be the direction of the force it would 
exert on a positive test charge. The electric field is radially outward from a 
positive charge and radially in toward a negative point charge. This is 
illustrated in the following examples.

Colinear charges:

      A       |       B
------+-------o-------+------------+---
    -0.02m    ^      0.02m        0.1m
            -5.5μC                2.5μC

EA = k(5.5μC)/(.02)2 - k(2.5μC)/(.12 m)2

The negative charge term is positive because the electric field at A 
is pointing in the positive x direction. While the positive charges term 
is negative because the electric field is pointing in the negative 
x direction. 

 EB = -k(5.5μC)/(.02)2 - k(2.5)/(.08)2

The negative charge term is negative because the electric field at B
is pointing in the negative x direction. While the positive charges 
term is negative because the electric field is pointing in the negative 
x direction.

Charges at corner of a square:

  A               B
 -2q  <-- d --> -3q
    . α
   ^  .
   |    .
   d      .
   |        .
   v          .
  +q            -4q
  D              C

  Force on A:

  x: FAB + FACcosα
  y: FAD + FACsinα

  B is pushing A to the left.
  D is pulling A downwards.
  Cx is pushing A upwards
  Cy is pushing A to the left.

Infinite line of charge:

        Ez
       /|
      / |   λ = charge per unit length
   r /  | z
    / α |
 ----------------------------  x->
-a                           b

dEz = (kλdx/r2)cosα = (kλdx/r2)(z/r)
         b          
    = kλz∫dx/(z2 + x2)3/2
         a

    = (kλ/z)[b/(z2 + b2) + a/(z2 + b2)]
 
As a and b -> ∞ E = 2kλ/z = λ/2πzε0

Charged ring:
    
      Ez
     /|
    / |
  r/  |z   
  /   |
 / θ  |
 ------------ Q
   R

Ez = kQcosθ/r2 = kQz/r3 = kQz/(z2 + R2)3/2

The max field is when dE/dz = 0 => z = R/√2

Charged disc:

          R
Ez = 2πkzσ∫R'dR'/(z2 + R'2)3/2
          0
   = 2πkσ[1 - z/(z2 + R2)]

Gauss's Law
-----------

The total electric flux ,φ, out of a closed surface is equal to the charge enclosed 
by the surface divided by the permittivity.   In integral form:

∫E.dA = φ = q/ε0

φ is perpendicular to A.  If not then φ = EAcosθ (dot product)

Gauss's law is an alternative to Coulomb's law for calculating the electric field 
due to a given distribution of charge.

Point charge:

  q ---->x  Ex = kq/r2  Coulomb
      r 

            φ = EA = q/ε0
            E = q/ε0A 
              = q/4πr2ε0 
              = kq/r2  

Infinite line of charge:

φ = EA = q/ε0 = λl/ε0   λ = charge per unit length
E = λl/2πε0rl 
  = λ/2πε0r 

Infinite sheet of charge (2 surfaces):

                      ^ E
                      |
                 ------------ A    φ = E2A = σA/ε0 
                      |
                      v E          E = σ/2ε

Electric Field Inside a Conductor
---------------------------------
In steady state the electric field inside a conductor must be 0 otherwise
charges would move around until they find an arrangement that makes the
electric field equal to 0 in the interior.  Any net charge must therefore
reside on the surface.  

         ^ E
         |
    ---------- A
   |   E = 0  |
    ----------

The outward flux from the surface is EA     

The charge enclosed = σA/ε0 where σ is the charge per unit area

Therefore, 

σA/ε0 = EA so E = σ/ε0


Ex 1.

Suppose that a steam of negatively charged particles is blown through a circular 
pipe with radius r = 5.0cm. Assume that the charge is spread throughout the volume
of particles in the pipe so that the charge density ρ is constant. (1) Find
the formula for the electric field at the pipe surface. (2)  Find the formula for
the field at a distance r1 from the pipe surface.

(1)
EA = q/ε0
q = πr2lρ
A = 2πrl

=> E = ρr/2ε0

(2) 
EA = q/ε0
q = πr2lρ
A = 2πr1l 

=> E = πr2lρ/2πr1ε0l = r2ρ/2r1ε0                         

Now,

ρ = q/πr2l

E = r2(q/πr2l)/2r1ε0 = (q/πl)/2r1ε0 = (q/l)/2πr1ε0 = λ/2πr1ε0

where λ is the charge per unit length.  This equation is the same as the 
one for the field due to an infinite line of charge.


Electric Potential Energy and Electric Potential (Voltage)
----------------------------------------------------------
Electric Potential energy, U, can be defined as the capacity for doing work 
which arises from position or configuration.

U = force * distance = kQq/r = qE.r = qErcosθ

In terms of calculus:

ΔU = ∫F.dr = ∫Frcosθ

     x
 A q--- B
   \θ  | 
  h \  | y
     \ | 
      \|
       C

UAB = qEx
UBC = 0
UAC = qEhcosθ = x

Therefore, the trajectory doesn't matter - only displacment in the direction 
of the field is counted.

If a positive charge moves in the same direction as the field then U will fall.
(the field does work on the charge). If it moves against the field, U will rise
(an external agent must do work on the charge).  The opposite is true for a
negative charge in the same field.

The voltage difference (aka electrical potential), V, is defined as the work done 
per unit charge against an electric field.
 
V = U/q = kQq/rq = kQ/r = (kQ/r2)r = Er

Therefore,

E = V/r and  U = qV

In terms of calculus:

ΔV = dW/q = (1/q)∫F.ds = ∫E.ds = ∫Ecosθds

E = -dV/dx

Ex 1. 

Electron in electric field.  How fast is it moving after distance s?

F = qE = ma

=> a = qE/m

v2 = 2as  assume initial velocity = 0

Alternatively, if we were given the potential difference, V

qV/s = ma

=> a = qV/ms

v2 =  2as  

   = (2qV/ms)s

   = 2qV/m

Note:  The final speed of the electron is entirely determined by its charge and the 
potential difference through which it is accelerated. 

We could also use the conservation of energy to solve the problem.

U = qV = (1/2)mv2

v2 = 2qV/m


Ex 2.  Dipole. Find points where electric potential is a certain +V
                           
   -Q                                             +Q
   -+----------------------|-------------+---------+--------+------->x
   -d                      O             A        +d        B

The electric potential will look something like:

               ||
              /  \
        -d   /    \
    --------0--------------
      \    /  +d
       \  /
        ||

There will be 2 points, A and B, that satisfy this condition.
  
VA = KQ/(x + d) - KQ/(d - x)

    = KQ{1/(x + d) - 1/(d - x)}

    = KQ{((x + d) - (d - x)/(x + d)(d - x)}

VA/KQ = 2x/(d2 - x2)

solve for x.

VB = KQ/(x + d) - KQ/(x - d)

    = KQ{1/(x + d) - 1/(x - d)}

    = KQ{((x + d) - (x - d)/(x + d)(x - d)}

VB/KQ = 2d/(x2 - d2)

solve for x.

 

Copyright © 2023 Redshift Academy

This website built by me using the Codeigniter MVC framework