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

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

2. 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

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 = 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)

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 = Fd/2 cos Energy of dipole U = – pE 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!!