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Physics 2102 Jonathan Dowling. Physics 2102 Lecture 6. Electric Potential II. –Q. +Q. U B. Example. Positive and negative charges of equal magnitude Q are held in a circle of radius R. 1. What is the electric potential at the center of each circle? V A = V B = V C =

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Physics 2102 lecture 6

Physics 2102

Jonathan Dowling

Physics 2102 Lecture 6

Electric Potential II






Positive and negative charges of equal magnitude Q are held in a circle of radius R.

1. What is the electric potential at the center of each circle?

  • VA =

  • VB =

  • VC =

    2. Draw an arrow representing the approximate direction of the electric field at the center of each circle.

    3. Which system has the highest electricpotential energy?




Electric potential of a dipole on axis





Electric Potential of a Dipole (on axis)

What is V at a point at an axial distance r away from the midpoint of a dipole (on side of positive charge)?

Far away, when r >> a:


Electric potential on perpendicular bisector of dipole


U= QoV=Qo(–Q/d+Q/d)=0 

Electric Potential on Perpendicular Bisector of Dipole

You bring a charge of Qo = –3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do:

  • Positive?

  • Negative?

  • Zero?






Continuous charge distributions
Continuous Charge Distributions

  • Divide the charge distribution into differential elements

  • Write down an expression for potential from a typical element — treat as point charge

  • Integrate!

  • Simple example: circular rod of radius r, total charge Q; find V at center.



Potential of continuous charge distribution example





Potential of Continuous Charge Distribution: Example

  • Uniformly charged rod

  • Total charge Q

  • Length L

  • What is V at position P shown?


Electric field potential a simple relationship
Electric Field & Potential: A Simple Relationship!

Focus only on a simple case:

electric field that points along +x axis but whose magnitude varies with x.

Notice the following:

  • Point charge:

    • E = kQ/r2

    • V = kQ/r

  • Dipole (far away):

    • E ~ kp/r3

    • V ~ kp/r2

  • E is given by a DERIVATIVE of V!

  • Of course!

  • Note:

  • MINUS sign!

  • Units for E -- VOLTS/METER (V/m)

Physics 2102 lecture 6 1340206

Electric Field & Potential: Example Simple Relationship!














  • Hollow metal sphere of radius R has a charge +q

  • Which of the following is the electric potential V as a function of distance r from center of sphere?

Physics 2102 lecture 6 1340206

Electric Field & Potential: Example Simple Relationship!



  • Outside the sphere:

  • Replace by point charge!

  • Inside the sphere:

  • E =0 (Gauss’ Law)

  • E = –dV/dr = 0 IFF V=constant


Equipotentials and conductors
Equipotentials and Conductors Simple Relationship!

  • Conducting surfaces are EQUIPOTENTIALs

  • At surface of conductor, E is normal to surface

  • Hence, no work needed to move a charge from one point on a conductor surface to another

  • Equipotentials are normal to E, so they follow the shape of the conductor near the surface.

Conductors change the field around them

An uncharged conductor: Simple Relationship!

A uniform electric field:

An uncharged conductor in the initially uniform electric field:

Conductors change the field around them!

Sharp conductors
“Sharp”conductors Simple Relationship!

  • Charge density is higher at conductor surfaces that have small radius of curvature

  • E = s/e0 for a conductor, hence STRONGER electric fields at sharply curved surfaces!

  • Used for attracting or getting rid of charge:

    • lightning rods

    • Van de Graaf -- metal brush transfers charge from rubber belt

    • Mars pathfinder mission -- tungsten points used to get rid of accumulated charge on rover (electric breakdown on Mars occurs at ~100 V/m)


Summary: Simple Relationship!

  • Electric potential: work needed to bring +1C from infinity; units = V

  • Electric potential uniquely defined for every point in space -- independent of path!

  • Electric potential is a scalar -- add contributions from individual point charges

  • We calculated the electric potential produced by a single charge: V=kq/r, and by continuous charge distributions : V= kdq/r

  • Electric field and electric potential: E= -dV/dx

  • Electric potential energy: work used to build the system, charge by charge. Use W=qV for each charge.

  • Conductors: the charges move to make their surface equipotentials.

  • Charge density and electric field are higher on sharp points of conductors.