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Chapter 15. Electric Forces and Electric Fields. A Question.

Chapter 15

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Chapter 15

Electric Forces and

Electric Fields

A Question

Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere C is uncharged. Sphere C is first touched to A, then to B, and finally removed. As a result, the electrostatic force between A and B, which was originally F, becomes:

A. F/2B. F/4C. 3F/8D. F/16E. 0

In Fig. at right, the particles have charges q1 = -q2 = 300 nC and q3 = -q4 = 200 nC, and distance a = 4.0 cm.

What is the direction of the force on particle 4?

A. B.

C. D.

[-0.344N, -0.218N], 212.38º

TALP: Take a look Problem

- Force on the center charge = 6kq2/d2
- Direction: -x, 180º,

d

- q1 = q4 = Q; q2 = q3 = q
- what is Q/q so that the force on 1 is zero
- All charges samesign. There is no solution.
- Q and q different sign. There is a possible solution.
- Is the answer the same if the force on 3 is to be zero

- What is the force on the central chlorine ion?
- Suppose that we put in an electron that sits on a cesium site and neutralizes it, which way does the Chlorine move?

- When no net motion of charge occurs within a conductor, the conductor is said to be in electrostatic equilibrium
- An isolated conductor has the following properties:
- The electric field is zero everywhere inside the conducting material
- Any excess charge on an isolated conductor resides entirely on its surface

- The electric field just outside a charged conductor is perpendicular to the conductor’s surface
- On an irregularly shaped conductor, the charge accumulates at locations where the radius of curvature of the surface is smallest (that is, at sharp points)

- The electric field is zero everywhere inside the conducting material
- Consider if this were not true
- If there were an electric field inside the conductor, the free charge there would move and there would be a flow of charge
- If there were a movement of charge, the conductor would not be in equilibrium

- Consider if this were not true

- Any excess charge on an isolated conductor resides entirely on its surface
- A direct result of the 1/r2 repulsion between like charges in Coulomb’s Law
- If some excess of charge could be placed inside the conductor, the repulsive forces would push them as far apart as possible, causing them to migrate to the surface

- The electric field just outside a charged conductor is perpendicular to the conductor’s surface
- Consider what would happen it this was not true
- The component along the surface would cause the charge to move
- It would not be in equilibrium

- On an irregularly shaped conductor, the charge accumulates at locations where the radius of curvature of the surface is smallest (that is, at sharp points)

- Any excess charge moves to its surface
- The charges move apart until an equilibrium is achieved
- The amount of charge per unit area is greater at the flat end
- The forces from the charges at the sharp end produce a larger resultant force away from the surface
- Why a lightning rod works

- Faraday’s Ice-Pail Experiment
- Concluded a charged object suspended inside a metal container causes a rearrangement of charge on the container in such a manner that the sign of the charge on the inside surface of the container is opposite the sign of the charge on the suspended object

- Millikan Oil-Drop Experiment
- Measured the elementary charge, e
- Found every charge had an integral multiple of e
- q = n e

- An electrostatic generator designed and built by Robert J. Van de Graaff in 1929
- Charge is transferred to the dome by means of a rotating belt
- Eventually an electrostatic discharge takes place

- Field lines penetrating an area A perpendicular to the field
- The product of EA is the flux, Φ
- In general:
- ΦE = E A cos θ

- ΦE = E A cos θ
- The perpendicular to the area A is at an angle θ to the field
- When the area is constructed such that a closed surface is formed, use the convention that flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive

- Gauss’ Law states that the electric flux through any closed surface is equal to the net charge Q inside the surface divided by εo
- εo is the permittivity of free space and equals 8.85 x 10-12 C2/Nm2
- The area in Φ is an imaginary surface, a Gaussian surface, it does not have to coincide with the surface of a physical object

- The calculation of the field outside the shell is identical to that of a point charge
- The electric field inside the shell is zero

- No force/field from the charge outside
- On inside circle E.4πr2=q/εo Coulomb’s law
- Far outside, E = k 5q/r2

P

+

+

+

+

+

+

- The electric field in a metal is zero
- Charge +q on inside surface.
- Because it is neutral, outside surface must be -q.

- Use a cylindrical Gaussian surface
- The flux through the ends is EA, there is no field through the curved part of the surface
- The total charge is Q = σA
- Note, the field is uniform

- The field must be perpendicular to the sheet
- The field is directed either toward or away from the sheet

- The device consists of plates of positive and negative charge
- The total electric field between the plates is given by
- The field outside the plates is zero

- Charges and forces on them
- Electric fields
- Gauss’ theorem
- Field near a plate
- Shells and wires
- Two plates