1. TOPIC 2 Electric Fields www.cbooth.staff.shef.ac.uk/phy101E&M/

2. Fields & Forces Coulomb’s law Q r q How does q “feel” effect of Q?
Q modifies the surrounding space.
Sets up electrostatic field E
Force on charge q is F = q E
E due to point charge Q is

3. Electrostatic Field Lines
Like charges Unlike charges
Field lines:
Start on positive charge, end on negative
Number proportional to charge
Strength of field = density of field lines
Direction of force at point = tangent to field line.

4. Conductors Charges flow in response to a field
Equilibrium  no net field within a conductor
Free charges only exist on surface
Consider components of field at surface:
Charge flows until E// = 0
 ETot is always perpendicular to surface of conductor

5. Continuous Charge Distributions
Divide into charge elements dq
Use superposition
In practice, express dq in terms of position r
 Use charge density
3D dq =  dV dV = element of volume
2D surface dq =  dA dA = element of area
1D line dq =  d d = element of length

6. Example 1
A rod of length L carries a charge Q distributed uniformly along its length. If it is centred on the origin and oriented along the y-axis, what is the resulting electric field at points on the x-axis?
Solution available on web page
Example 2
A charge Q is uniformly distributed along the circumference of a thin ring of radius R. What is the electric field at points along the axis of the ring?
For next lecture: revise binomial theorem.

7. Electric Dipoles
Pair of equal & opposite charges, Q & –Q, separated by distance d
Dipole moment (vector) p = Q d (direction is from negative to positive charge)
Total charge is zero, but still produces and experiences electric fields
In uniform electric field, dipole experiences a torque (though no net force)

8. Pair of equal & opposite forces F = QE
Perpendicular separation between lines of forces = d sin 
Torque  = F d sin  = Q E d sin  = p E sin 
As vector,  = p  E
i.e. torque acting about centre of dipole, tending to rotate it to align with electric field

9. Would have to do work to rotate dipole away from aligned position – stored as potential energy.
Dipole does work (loses energy) rotating towards aligned position.
Define zero of potential energy when dipole is perpendicular to field –  = 90°.
Rotating to position shown, each charge does work: work =forcedistance = F d/2 cos
Energy of dipole U = – p E cos  = – p.E

10. Example 1
What is the electric field at points on the x-axis due to a dipole formed by a charge Q at x = a/2 and a charge –Q at x = –a/2 , for values of x >> a?
Example 2
Two dipoles, with the same charge and separation as above, are placed parallel and a distance  apart: (a) parallel to the line of the dipoles (b) perpendicular to the line of the dipoles. In which case is the force between the dipoles greatest? Part (b) is HARD!!