Calculating Electric Fields and Forces using Coulomb's Law and Gauss' Law, Lecture notes of Electromagnetic Engineering

Download Calculating Electric Fields and Forces using Coulomb's Law and Gauss' Law and more Lecture notes Electromagnetic Engineering in PDF only on Docsity!

Welcome…

…to Electrostatics

Outline

1. Coulomb’s Law

2. The Electric Field

- Examples

3. Gauss Law

- Examples

4. Conductors in Electric Field

To make this into a “really good” starting equation I should specify “repulsive for like,” but that makes it too wordy. You’ll just have to remember how to find the direction.

1 2 2 12

q q

F k ,

12 r

 attractive for unlike

Force is a vector quantity. The equation on the previous slide gives the magnitude of the force. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/4 0 , where  0 =8.85x10-12^ C^2 /N·m^2.

Remember, a vector has a magnitude and a direction.

One could write Coulomb’s Law like this…

The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed, r 12 is the distance between the centers of the spheres.

If more than one charge is involved, the net force is the vector sum of all forces (superposition). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!).

r 12

Solving Problems Involving Coulomb’s Law and

Vectors

x

y

Q 2 =+50C

Q 3 =+65C

Q 1 =-86C 52 cm

30 cm =30º

Example: Calculate the net electrostatic force on charge Q 3 due to the charges Q 1 and Q 2.

Step 0: Think!

This is a Coulomb’s Law problem (all we have to work with, so far).

We only want the forces on Q 3.

Forces are additive, so we can calculate F 32 and F 31 and add the two.

If we do our vector addition using components, we must resolve our forces into their x- and y-components.

1 2 2 12

q q

F k

12 r

“Do I have to put in the absolute value signs?”

x

y

Q 2 =+50C

Q 3 =+65C

Q 1 =-86C 52 cm

30 cm =30º

F 31

F 32

Step 2: Starting Equation

3 2 2 32

Q Q

F k ,

32 r

repulsive

3 2 2 32

Q Q

F k

32, y r

F 0

32, x

 (from diagram)

F32,y = 330 N and F32,x = 0 N.

x

y

Q 2 =+50C

Q 3 =+65C

Q 1 =-86C 52 cm

r^32

=30 cm =30º

F 31

F 32

Step 3: Replace Generic Quantities by Specifics

F3x = F31,x + F32,x = 120 N + 0 N = 120 N

F3y = F31,y + F32,y = -70 N + 330 N = 260 N

You know how to calculate the magnitude F 3 and the angle between F 3 and the x-axis.

F 3

The net force is the vector sum of all the forces on Q 3.

x

y

Q 2 =+50C

Q 3 =+65C

Q 1 =-86C 52 cm

30 cm =30º

F 31

F 32

Step 3: Complete the Math

A sample Coulomb’s law calculation using three point charges, which can be extended upto infinite number of point charges in the form of continuous charge distributions on any type of 1D,2D and 3D objects.

How do you apply Coulomb’s law to objects that contain distributions of charges?

We’ll use another tool to do that…

WHAT WE HAVE DONE

QUESTION?

Faraday, beginning in the 1830's, was the leader in developing the idea of the electric field. Here's the idea:

 A charged particle emanates a "field" into all space.

 Another charged particle senses the field, and “knows” that the first one is there.

like charges repel

unlike charges attract

F 12

F 21

F 31

F 13

We define the electric field by the force it exerts on a test charge q 0 :

0 0

F

E =

q

By convention the direction of the electric field is the direction of the force exerted on a POSITIVE test charge. The absence of absolute value signs around q 0 means you must include the sign of q 0 in your work.

If the test charge is "too big" it perturbs the electric field, so the “correct” definition is

0

0 q (^0 )

F

E = lim

 q

Any time you know the electric field, you can use this equation to calculate the force on a charged particle in that electric field.F = qE

The Electric Field Due to a Point Charge

Coulomb's law says

... which tells us the electric field due to a point charge q is

1 2 2 12

q q F =k , (^12) r

q (^2)

q

E =k , away from +

r

2

q

E=k

r

We define as a unit vector from the source point to the field point:

ˆr

source point

field point

ˆr The equation for the electric field of a point charge then becomes:

2

q E=k rˆ r

Motion of a Charged Particle

in a Uniform Electric Field

A charged particle in an electric field experiences a force, and if it is free to move, an acceleration.

---

• • • • • • • • • • + + +

• F

If the only force is due to the electric field, then

F^ ^ ma^ qE.

If E is constant, then a is constant, and you can use the equations of kinematics.

E