lecture20 electrostatics

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Electrostatics Solving problems involving stationary electric fields

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Lecture for Payap University General Science Course

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Page 1: Lecture20 electrostatics

Electrostatics

Solving problems involving stationary electric fields

Page 2: Lecture20 electrostatics

Electrostatics

Definition of Flux:Definition of Flux:•The amount of something (field, material or other physical The amount of something (field, material or other physical entity) passing through a surface.entity) passing through a surface.•Surface area can be represented as vector defined normal to Surface area can be represented as vector defined normal to the surface it is describingthe surface it is describing

Page 3: Lecture20 electrostatics

Electrostatics

Electric Flux: The amount of electric field passing The amount of electric field passing through a surface areathrough a surface area

A

E

(The angle is the angle between a normal to the surface and the electric field)

Page 4: Lecture20 electrostatics

Consider water source (spring). Suppose you enclose the spring with a closed surface, such as a sphere. If your water accumulates within the sphere, you can see that the total flow out of the sphere is equal to the rate at which the source is producing water. In the case of electric fields the source of the field is the charge. So we can now say that the sum of the sources within a closed surface is proportional to the total flux through that surface. This is known as Gauss' Law

The vacuum permittivity, also known as the electric constant is the constant of proportionality in this case

Gauss’ Law

Carl Friedrich Gauss 1777-1855

8.854 × 10−12 C2·N−1·m−2

Page 5: Lecture20 electrostatics

An imaginary closed surface created to enable An imaginary closed surface created to enable the application of Gauss’s Lawthe application of Gauss’s Law

What is the total flux through each surface?What is the total flux through each surface?

A Gaussian surface

Page 6: Lecture20 electrostatics

A Gaussian surface

A Gaussian surface that completely surrounds a point charge intercepts the same number of field lines regardless of its shape.

For a positive charge, the lines exit the surface; for a negative one they enter it.

Page 7: Lecture20 electrostatics

Gauss’s Law for Electric Fields

If a greater amount of charge is enclosed, more field lines cross the surface.

Page 8: Lecture20 electrostatics

Gauss’s Law for Electric Fields1. Surface 1 surrounds the

positive charge and has lines exiting it.

2. Surface 2 surrounds the negative charge and has lines entering it.

3. Surface 3 does not enclose any charge, and the same number of lines exit as enter.

4. Surface 4 encloses both charges; as they are equal in magnitude, the same number of lines exit the surface as enter it.

Page 9: Lecture20 electrostatics

Gauss’s Law for Electric FieldsThe net number of electric field lines passing through an imaginary closed surface is proportional to the amount of net charge enclosed within that surface.

This can be used to show that excess charge on a conductor must reside on the surface:

E is 0 inside, so electric flux through Gaussian surface just inside conductor is 0, so no net charge is enclosed.

Page 10: Lecture20 electrostatics

Using Gauss’ LawCan we use the law to calculate electric field from a point charge?

This is equation for field from a point charge, so it works!

We know field from positive point charge points away from charge. Put Gaussian sphere around it, with radius r. What’s the flux?

Gauss:

So:

Page 11: Lecture20 electrostatics

Make a Gaussian pill box around a portion of it in the form of a cylinder. What’s flux through the cylinder?

•No field points through the ends (symmetry)

•Field points perpendicular to side of cylinder

Gauss:

So:

Using Gauss’ LawCan we use the law to calculate electric field from a lot of charges arranged along a line? Imagine an infinitely long wire with charge density =q/L (charge/length)

This is the field from a uniformly charged long wire

+++++

++

+

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Page 12: Lecture20 electrostatics

One last bit on electric fieldsLet’s look at the strength of the electric force: electrons in the human brain if we took away all the protons. There are roughly 4.2 x 1026 electrons in your brain. Let’s assume the brain is a sphere with radius 10 cm, calculate force on one electron near the surface:

Putting in numbers, we get about 10 Newtons of force on one electron. Considering the mass of electron is only 9.1x10-

31kg, acceleration would be pretty severe: F= ma…

Page 13: Lecture20 electrostatics

One last bit on electric fieldsLet’s assume the electrons are uniformly distributed on the surface of the brain (as if the brain were a conductor). What’s the force per unit area? (F/A is called pressure)

When I stand on the ground, what pressure do I exert? Roughly,

So pressure from un-neutralized charge in the brain = 30,000,000,000,000 men standing on your head