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

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

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

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

- 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.

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

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

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

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 =forcedistance = Fd/2 cos

Energy of dipole U = – pE cos = – p.E

Example 1 position – stored as potential energy.

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!!

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