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

1 / 10

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

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

PowerPoint Slideshow about ' TOPIC 2 Electric Fields cbooth.staff.shef.ac.uk/phy101E&M/' - xia

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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

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

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!!