this week week of january 18 chemistry 1310 midterm · → an electric field can exist inside an...

This WeekExperiment 1: Equipotential Lines

Week of January 18Tutorial and Test 1 on chapter 18

Chemistry 1310 MidtermIs NOT at the same time as the PHYS 1030 midterm!

Aurora: CHEM1310, Thursday, March 4, 7-9 pm

58Monday, January 11, 2010

• Conductor contains charges that are free to move (electrons)

• At equilibrium, the charges are at rest

Electrical Conductor

E = 0

! there can be no electric field inside the conductor, otherwise the charges would be moving under the influence of the field and would not be in equilibrium.

The electric fields due to the rod and the charges on the surface cancel each other out inside the conductor


E = 0

The extra electrons are pushed to the surface of the conductor by the repulsive Coulomb forces between them.

As the electrons end up at equilibrium (i.e. at rest) there can be no electric field inside the conductor, otherwise the electrons would still be moving...

Put some extra electron charges inside the conductor

Electrical Conductor


E =q

�0A

Coulomb constant, k =1

4π�0

Uniform field inside a parallel plate capacitor

+q -q

Area A

�E

Uniform spacing of field

lines means constant

electric field


At equilibrium, electric field lines are made to hit the conductor at right angles – charges move around the surface, distorting the electric field outside, until they reach an equilibrium in which there is no longer a field parallel to the surface to move them further.

Uniform spacing of field lines means constant electric field

E = 0

Distortion of field lines due to charges on surface of conductor

–e E=0


Equipotential Surface As the electric field lines are at right angles to the conducting surface, no work is done in moving an electron charge around the surface

! the electron has constant potential energy on the surface

! the surface is an “equipotential surface” (more in chapter 19)

– equipotential surfaces are at right angles to electric field lines

Electric potentials are measured in volts (V).

–eE = 0

�E


Experiment 1 : Equipotential Lines

Sketch out equipotential lines by using a galvanometer to find places that lie at the same potential (voltage).

– one probe of the galvanometer is fixed in position, the other is moved around to locate points at the same potential as the galvanometer - no current flows through the galvanometer.

– all of these points lie on an equipotential line.

Electric field lines are sketched in so that they are always at right angles to the equipotential lines.


Equipotentials and electric field

V1 volts

V2 volts

90º

90º 90º

90º


Equipotentials and Field Lines

V = 0

EE

V

V


E = 0

E = 0

Scooping out material where there is no electric fieldresults in a cavity with no electric field

Interior of the cavity is shielded from outside fields – basis of “Faraday cage”– example, shielding of electronic circuits (in a closed metal box) from stray electric fields


Clicker Question 18: The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?

A) AB) BC) CD) D

Answer: D) D


The number of field lines leaving +q is equal to the number hitting the inner surface of the conductor...

! there must be an induced charge of –q on the inner surface of the conductor (the field lines point toward the inner surface, so the induced charge must be negative).

As the outer surface has a charge +q, it must have the same number of lines leaving the surface ! from the outside, the conductor has no effect on the electric field.

–q

+q

As the conductor is electrically neutral (has zero net charge), there must also be an induced charge +q on the outer surface of the conductor.

E = 0

Charge +q placed inside an uncharged hollow conductor

Text


Clicker QuestionA metal sphere with a hollow centre has a charge of +2q, initially all on the outer surface.

An additional charge +q is placed in the hollow centre of the sphere. The charge on the outer surface of the sphere becomes:

A) -qB) 0C) +qD) +2q E) +3q

+q Metal sphere

+2q

Hints: The number of field lines is proportional to charge. What is the charge on the inner surface of the conductor? Charge is conserved.

Answer: E) +3q


-q

Qouter = +3q• All lines of force from +q at centre end up on inner surface of sphere! charge -q on inner surface of sphere

• Charge conserved: total charge on sphere is:

-q + Qouter = 2q

So Qouter = +3q


Inside an Electrical Insulator

Electric charges cannot move, so:

→ charges do not move around to reduce the electric field to zero

→ an electric field can exist inside an insulator


Electric FluxA measure of the number of electric field lines passing through a surface; the greater the number, the greater the electric field.

Gaussian surface: a closed surface

Sum over the Gaussian surface

Component of field at right angles to

the surface


Prob. 18.57/49

A vertical wall (5.9 m x 2.6 m) in a house faces due east. A uniform electric field has a magnitude of 160 N/C. This field is parallel to the ground and points 41o north of east. What is the electric flux through the wall?


True for all cases where q is the total charge enclosed by the Gaussian surface

The electric flux passing through the Gaussian surface is !E = EAA = surface area of the Gaussian sphere = 4πr2

Gauss’s Law

Choose a spherical Gaussian surface of radius r with a charge q at its centre.

E is the same all the way around the surface:

Gauss’s Law - one charge


+q

Uncharged spherical

conductor, E = 0 inside

Spherical Gaussian surface

The electric flux passing out through the Gaussian surface is:

!E = EA = 0, as E = 0 inside the conductor.

Gauss’s Law

So, the total charge inside the Gaussian sphere is zero!

–q

There must be a charge -q on the inner surface of the conductor...and +q on the outer surface as the conductor has zero net charge.From outside, looks like a charge q at the centre of the sphere.

E = 0

+q


Parallel Plate Capacitor+q –q

E

E

E

E = 0 inside the metal

In red: a cylindrical Gaussian tube. The area of the ends is "A. There are charges ±q on the plates of area A.

The only electric flux leaving or entering the tube is through the right hand end of area "A.

Gauss:

Charge "qArea "A

Total area = A

Plates of area A

The same all the way across the capacitor


Prob. 18.76/24: There are four charges, each of magnitude 2 !C. Two are positive and two are negative.

The charges are fixed to the corners of a 0.3 m square, one to a corner, in such a way that the net force on any charge is directed to the centre of the square.

Find the magnitude of the electrostatic force experienced by any charge.

Work out how to arrange the four charges so that the net force on each is toward the centre of the square –"implies there must be some symmetry in the arrangement of charges.


this week week of january 18 chemistry 1310 midterm · → an electric field can exist inside an...

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