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TOPIC 2 Electric Fields

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TOPIC 2 Electric Fields 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.

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fields forces
Fields & Forces
  • Coulomb’s law
  • Qrq
  • How does q “feel” effect of Q?
    • Q modifies the surrounding space.
    • Sets up electrostatic field E
    • Force on charge q is F = qE
    • E due to point charge Q is
electrostatic field lines
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.
  • 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
continuous charge distributions
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


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.

electric dipoles
Electric Dipoles

Pair of equal & opposite charges, Q & –Q, separated by distance d

Dipole moment (vector) p = Qd(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)


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


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 = Fd/2 cos

Energy of dipole U = – pE cos  = – p.E


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!!