1 TOPIC 2 Electric Fields

2 Coulomb’s law QrqQrq 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 Fields & Forces

3 Like chargesUnlike 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. Electrostatic Field Lines

4 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  E Tot is always perpendicular to surface of conductor Conductors

5 Divide into charge elements dq Use superposition In practice, express dq in terms of position r  Use charge density 3Ddq =  dVdV = element of volume 2D surfacedq =  dAdA = element of area 1D linedq = d d = element of length Continuous Charge Distributions

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? See handout Question 3. For next lecture: revise binomial theorem.

7 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) Electric Dipoles

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!!