Download Coulomb`s Law Script Slide One: Click Here Button

Coulomb’s Law Script
Slide One: Click Here Button
Charge and Electrical Force Slide: Charges exert forces on each other we have
called them attraction and repulsion so far. We know that opposite charges attract
and like charges repel.
Coulomb’s Law Slide: As an equation Coulomb’s Law is Force is equal to k times
charge 1 times charge 2 all divided by r squared. F is the force between charges 1
and 2. k is a constant of 9.0 x 109 Newtons meters squared divided by Coulomb’s
squared. Q1 is charge 1 and q2 is charge 2. If you calculate the force and it turns
out to be positive then you know it is a force of repulsion. If you calculate the force
and it turns out to be negative then it is a force of attraction.
Example Problem 2 Slide: A point charge of 3.8 µC is located 12.0 cm from
another point charge of 5.1 µC. What is the force on the 5.1 µC charge?
Example Problem 2 Solution: Using Coulomb’s Law we are going to calculate the
force on the 5.1 µC charge, but we know from mechanics that forces occur in pairs
so this is also the force on the 3.8 µC charge.
Coulomb’s Law states that force is equal to the constant, k, times charge 1 times
charge 2 divided by the radius squared.
So we set up the equation Force is equal to 9 x 10 9 Nm2/C2 times the first charge
3.8 x 10-6 C (we must convert from microCoulombs to Coulomb’s) times 5.1 x 10-6
C. Now we divide by the radial distance separating them 12.0 cm, but we have to
change centimeters to meters. We divide 12 by 100 and get 0.12 meters. So divide
by 0.12 m squared. That gives us an answer of 12. Now let’s look at the units. The
Coulomb’s cancel out and the meters squared cancel out so we are left with
Newtons for our unit so our answer is 12 Newtons. This is positive so that means it
is a force that is repelling them from each other.
Superposition Slide: Electrical forces are like other forces they are vector
quantities and must be added as vectors. If more than one force is acting on a
charged object then you must calculate the force from each charge acting on the
object then add up all the forces acting on the object using vector math. This is
known as the principle of superposition.
Multiple Forces on a Single Charge Slide: Here we have two forces acting on
the -2.5 µC charge. We want to find the net force acting on it.
Solution Slide: We must first determine the force from each charge acting on the
-2.5 µC charge. That means we have two forces acting on it. The 3.0 µC charge at
the bottom is attracting it downward and the other 3.0 µC charge is attracting it to
the right. Now let’s find the magnitude of these forces. F=9x109Nm2/C2 times 3.0 x
10 -6 C times -2.5 x 10-6 C divided by 0.10 m squared. This gives us a magnitude of
11.5 Newtons. The charges and the distance are the same so both forces are 8.1
Newtons. Now we have to add the two vectors together. We use Pythagorean’s
Theorem to do that so Fnet = the square root of 8.1 N squared plus 8.1 N squared.
That gives us a net force of 11.5 Newtons. The direction is 45 degrees below the x
axis or below the line connecting the -2.5 µC charge and the 3.0 µC charge to the
right of it.
Example Problem 3 Slide: A charge q1 = 4 µC is positioned at the origin. A
charge of q2 = 9 µC is positioned at x = 4 m. Where on the x-axis can a charge q3
be placed so that the force is zero?
Example Problem 3 Solution Slide: We need to decide where to put the third
charge here so that the net force on it is zero. That means the forces from the
other two charges must add up to equal zero or they must be equal and opposite.
That means that I will set the two forces equal to each other and solve for the
distance from the charge placed at the origin. Because both charges are
measured in micro coulombs, we can cancel out the exponent 10 to the
negative sixth and use the numbers four and nine in our equation.
So that means (kq1q3)/x2 = (kq2q3)/ (4-x)2 . I set the distance from the origin equal
to x and the other distance is 4-x. Now we use the quadratic equation to solve for x
and plug in the values for the other variables. The values 1.6 and negative 8
are arrived at using the quadratic equation. The value negative 8 is not
valid because it is not between the two other charges. That means it is 2.4 m
from the other charge or 12 m from the other charge. Both are valid answers.
Triangle Practice Problem Slide: Now let’s try something more complicated.
What is the force on a charge of 1.0 µC charge placed at point P? The height of the
triangle is 1.1 m and the sides of the equilateral triangle are 1.0 m.
Solution Slide:We first find the force from each charge on the 1.0 microcoulomb
charge placed at point p. We use the equation force equals kq1q2 divided by r2. For
the first charge starting to the right of point p, I will call this F1, F1 = F =
1
9
-6
-6
2
9x10 (2.0 x 10 )(1.0 x 10 )/(0.5 ) So F = 0.072 N. Now let’s look at the next
1
9
-6
-6
2
charge F2 at the top of the triangle. F = 9x10 (2.0 x 10 )(1.0 x 10 )/(1.1 ) So F
2
9
-6
-6
2
= 0.0149 N. Now let’s do F3. F = 9x10 (4.0 x 10 )(1.0 x 10 )/(1.0 )
3
So F = 0.036 N
3
2
Solution Continued Slide: Now we look at the direction of each vector. F1 and F3
point to the left at point p’s charge. So we add those two together getting 0.108 N.
F2 is pointing down at P so we add these two forces together using Pythagorean’s
theorem. The net force is equal to the square root of 0.108 N squared plus 0.0149
N squared. This gives us a net force of 0.109 Newtons. We do the inverse tangent
of 0.0149 divided by 0.108 and we get the angle with respect to the negative x
axis. This angle is 7.9 degrees.