>> In this video,
we're gonna take a Closer Look
at advanced applications
of Coulomb's Law.
Remember that Coulomb's Law
says the magnitude
of the electric force
between two charges,
q1 and q2,
equals a constant number,
we'll call it k,
times the absolute value
of the charge, q1,
times the charge q2,
all divided by the distance
between q1 and q2.
We'll call it r-squared.
The constant number, k,
equals 9 times 10
to the ninth newton
meters squared
per coulomb squared.
What if we wanna know the
electric force on a charge
due to multiple other charges?
In this example,
q1, q2 and q3 are all
in a line on the x axis.
And we're told the charge
of each one.
Knowing electric force has
both magnitude and direction,
what is the net electric force
on the middle charge, q2?
To find out, we need to compute
the electric force on q2
due to q1,
the electric force on q2
due to q3,
and then add them together
to solve for the total
electric force.
Using Coulomb's Law,
the electric force magnitude
of q1 and q2 equals
k times the absolute value
of q1 times q2,
divided by the distance
between q1 and q2 r-squared.
Plugging in 9 times 10
to the ninth newton
meters squared
per coulombs squared for k,
2 times 10 to the negative tenth
coulombs for q1,
negative 4.5 times 10
to the negative thirteenth
coulombs for q2,
and 1.50 meters for r,
we find that the electric force
for magnitude of q1 on q2
is 3.6 times 10 to the
negative thirteenth newtons.
Since q1 and q2
have opposite signs,
they attract one another.
That means q1 pulls q2
in the negative x direction.
The next step is to figure out
the electric force magnitude
of q3 on q2.
Plugging in for k,
9 times 10 to the ninth
newton meters squared,
divided by coulombs squared,
negative 5 times 10 to
the negative ninth coulombs
for q3,
negative 4.5 times 10 to the
negative thirteenth coulombs
for q2, and 2.0 meters for r.
We find the electric force
magnitude of q3 on q2
is 5.1 times 10 to the
negative twelfth newtons.
Since q3 and q2
have the same sign,
they push one another away.
This means q3 pushes q2
in the negative x direction.
The last step to finding
the net electric force on q2
is to add these
two forces together.
Since electric force
is a vector quantity,
and the forces both point
in the negative x direction,
we find the net magnitude
by adding the individual
magnitudes together.
The force of q1 on q2,
negative 3.6 times 10 to
the negative thirteenth newtons,
plus the force of q3 on q2,
negative 5.1 times 10 to
the negative twelfth newtons,
equals negative 5.5
times 10 to the negative
twelfth newtons,
or 5.5 times 10 to
the negative twelfth newtons,
pointing in the
negative x direction.
Now let's study another
example where, this time,
three charges are arranged
at the corners of
an equilateral triangle,
with sides 75 centimeters long.
Once again, we're given
the charges of q1, q2 and q3.
We want to know what is
the net electric force on q2?
The charge at the top
of the triangle.
The net electric force on q2
will be the sum of the forces
q1 and q3 exert on q2.
Let's start with q1.
We can find the magnitude
of the electric force
q1 exerts on q2
using Coulomb's Law.
The force equals k,
9 times 10 to the ninth
newton meters squared
per coulomb squared,
times absolute value of q1,
2 times 10 to the negative
tenth coulombs, times q2,
negative 4.5 times 10 to
the negative twelfth coulombs,
all divided by the distance
between q1 and q2, r,
or 0.75 meters squared.
Plugging in these numbers
gives us 1.4 times 10 to
the negative eleventh newtons.
Now, in what direction
does the force point?
Since q1 and q2
have opposite charges,
we know the force between
them is attractive.
Since we will add this force
to another one later on,
what we'd really like to know
is the amount of this force
when it is broken up
into its components
in the x and y directions.
We can use the fact that
drawing a vertical line
from q2 down to the x axis
creates a 30-60-90 triangle.
What we want to find is
the x and y components of
the electrical force of q1
acting on q2, which we call
F sub x and F sub y.
The fact that we have
a special right triangle
lets us solve for F sub x
and F sub y using ratios.
The two triangles in
this diagram are similar.
That means the ratio of
the total force of q1 on q2 to
the x component of that force
is equal to two divided by one.
Multiplying both sides
of the equation by F sub x
and dividing both sides by two
shows that F sub x equals
1.4 times 10 to the
negative eleventh newtons
divided by two,
which equals 7.0 times 10
to the negative twelfth newtons
in a negative x direction.
Now, let's solve for F sub y
a similar way using a ratio.
Going back to our
similar triangles,
the total electric force of q1
on q2, divided by F sub y,
equals two divided by
the square root of three.
To solve for F sub y,
first we multiply both sides
by F sub y,
then multiply both sides by
the square root of three
divided by two.
The force of q1 on q2
in the negative y direction
equals 1.4 times 10
to the negative eleventh,
times the square root
of three over two,
which equals 1.2 times 10 to
the negative eleventh newtons
in the negative y direction.
We've found the x and y
components of the force
of q1 on q2.
We're halfway there to solving
for the total electric force
on q2 in this charge
distribution.
The third charge, q3,
also exerts a force on q2.
Just like with q1,
we'll solve for the magnitude
and direction of that force.
By Coulomb's Law,
the magnitude of the force
of q3 on q2 is k.
9 times 10 to the ninth
newton meters squared
per coulombs squared,
times the absolute value of q2.
Negative 4.5 times 10 to
the negative twelfth coulombs,
times q3.
Negative 5 times 10 to
the negative tenth coulombs,
divided by the distance
between the charges, r-23,
or 0.75 meters squared.
This equals 3.6 times 10 to
the negative eleventh newtons.
That's the magnitude of
the electric force of q3
acting on q2.
Since q3 and q2 both
have negative charges,
they repel each other.
The force on q2
points away from q3.
Once again, we'll use
our 30-60-90 triangle
to find out the
force components.
We'll let F sub x be the
x component of the force,
and F sub y be the y component.
Solving first for F sub x,
we see from the similar
triangles that the total
force of q3 on q1,
3.6 times 10 to the negative
eleventh newtons,
divided by F sub x,
equals two divided by one.
Multiplying both sides
by F sub x and dividing
both sides by two,
we find that F sub x
equals 3.6 times 10
to the negative eleventh,
divided by two,
which is 1.8 times 10 to
the negative eleventh newtons
in the negative x direction.
To solve for F sub y,
we'll use the fact that
3.6 times 10 to the negative
eleventh newtons,
divided by F sub y equals 2,
divided by the
square root of 3.
Multiplying both sides
by F sub y, and then by
the square root of three
divided by two,
shows us that F sub y
equals 3.6 times 10 to the
negative eleventh newtons,
times the square root of three,
divided by two,
which equals 3.12 times 10 to
the negative eleventh newtons
in the positive y direction.
Now we are ready to solve
for the total magnitude
and direction of the
electric force on q2
from q1 and q3.
We'll use this table to
add up the x and y components.
To find the total force
on q2 in the x direction,
we add up the x components
of the force of q1 on q2,
and q3 on q2.
This equals 2.5 times 10 to
the negative eleventh newtons
in the negative x direction.
To find the total force on q2
in the y direction,
we add up the y components
of the force of q1 on q2,
and of q3 on q2.
This equals 1.92 times 10 to
the negative eleventh newtons
in the positive y direction.
Using the Pythagorean theorem,
we can take the square root
of the sum of the squares
of these components
to find the magnitude
of the net force on q2.
The square root of
negative 2.5 times 10
to the negative eleventh
newtons squared,
plus 1.92 times 10 to
the negative eleventh squared
equals 3.2 times 10 to the
negative eleventh newtons.
That's the magnitude of
the total force on q2.
The direction of this net force
is found using
the tangent function.
The angle the force forms
with the positive x axis,
we'll call it theta,
equals the arctangent of
the sum of the y components
of the forces acting on q2,
which is 1.92 times 10 to
the negative eleventh newtons,
divided by the sum of the
x components of the forces
acting on q2,
which is negative 2.5 times 10
to the negative
eleventh newtons.
The result we find is
negative 37.5 degrees.
But since the angle lies
in the second quadrant,
we'll add 180 degrees
to that result.
180 degrees plus negative
37.5 degrees equal 142 degrees
with respect to
the positive x axis.
Let's do one last
Coulomb's Law exercise.
This time with four charges
arranged in a quadrilateral.
We're given the amount
of each charge,
and we want to figure out what
the total electric force is,
magnitude and direction,
on charge q3.
We'll need to know how far
from q3 the rest of
the charges are.
So let's solve for those
distances as the first step.
The distances in red are
the ones we want to figure out.
r-13, r-23, and r-43.
r-13 is already known.
It's 75 centimeters,
or 0.75 meters.
To figure out r-23,
let's focus on
the highlighted triangle.
First, we'll solve for
the angle marked as theta.
We can see that theta
plus 117 degrees
equals 180 degrees.
Subtracting 117 degrees
from both sides,
theta equals 180 degrees
minus 117 degrees,
or 63 degrees.
Now we can use the sine
trigonomic function
to solve for r-23.
Sine theta equals
the side opposite the angle,
divided by the hypotenuse.
Plugging in our values,
the sine of 63 degrees
equals 60 centimeters,
divided by r-23.
After we multiply both sides
by r-23 and divide both sides
by the sine of 63 degrees,
we see that r-23
equals 67 centimeters.
To figure out the last
unknown distance r-43,
we'll use our sine
function again.
In this case,
the sine of our angle,
28 degrees,
equals 31 centimeters
divided by r-43.
After we multiply both sides
by r-43 and divide both sides
by the sine of 28 degrees,
we see that r-43
equals 66 centimeters.
Now that we have
the distance between q3
and the other charges,
and the charge values
themselves,
we can use Coulomb's Law
to solve for the electric
force magnitude between
each pair of charges.
Remember that Coulomb's Law
says the electric force
magnitude between two charges,
q1 and q2,
equals the constant, k,
times the absolute value
of q1 times q2,
all divided by r,
the distance between
the charges squared.
This table summarizes what
we know about the scenario,
and what we want to solve for,
the electric force.
Starting with the first
row of values,
we plug these into Coulomb's Law
and find that the
electric force magnitude
between q1 and q3
is 6.4 times 10 to
the negative tenth newtons.
We fill that in on the table
and move onto the next row,
q2 and q3.
Plugging these values
into Coulomb's Law,
we find that the electric force
between q2 and q3
is 4.0 times 10 to the
negative ninth newtons.
Now for the last
pair of charges,
q4 and q3.
Coulomb's Law tells us that
the force between q4 and q3
is 4.1 times 10 to
the negative ninth newtons.
Now we know all the force
magnitudes and just need to
figure out their directions.
Here, the magnitude
and direction of the forces
on q3 are shown.
The direction of the arrows
depends on whether
each pair of charges
repels each other,
like q1 and q3,
or attracts each other,
like the other two pairs.
We want to solve for x and y
components of each of
these three vectors,
so we can add them all together
to find the resultant force
magnitude and direction.
Here's a zoomed in view of
the forces acting on q3,
along with the electric force
components we want to solve for.
To figure out the x and y
components of the forces,
we can use the sine
and cosine functions.
Since we know all
the angles involved,
all we'll need to do is use
sine and cosine of those angles
and multiply that
by the force magnitude.
The table shows what
we want to calculate.
Let's make all of our angles
with respect to
the positive x axis.
This will allow us
to consistently use
the cosine function
with our x components
and the sine function
with our y components.
Starting with the top row,
since the cosine
of zero degrees is one
and the sine
of zero degrees is zero,
the x component of
the force of q1 on q3
is 6.4 times 10 to
the negative tenth.
And the y component is zero.
For the x component of
the force from q2 on q3,
4.0 times 10
to the negative ninth,
times cosine of 117 degrees
equals negative 1.8 times 10
to the negative ninth newtons.
The y component equals
4.0 times 10 to the
negative ninth newtons,
times the sine of 117 degrees,
which is 3.6 times 10 to
the negative ninth newtons.
To find the x and y components
of the force of q4 on q3,
multiply the force magnitude,
4.1 times 10 to the
negative ninth newtons,
times the cosine and sine
of 62 degrees, respectively.
The x component equals
1.9 times 10 to the
negative ninth newtons.
And the y component equals
3.6 times 10 to the
negative ninth newtons.
Adding up all the x components
and all the y components,
we find the total resulting
x and y force components
acting on charge q3
to be 7.4 times 10 to
the negative tenth newtons
in the x direction.
And 7.2 times 10 to the
negative ninth newtons
in the y direction.
By the Pythagorean theorem,
the total electric force
acting on q3 squared
equals the sum of
the x components of force
acting on q3 squared,
plus the sum of the
y components of force
acting on q3 squared.
Taking the square root of
both sides of the equation
and plugging in the x and y
components of the force on q3,
we find that the total
electric force on q3 rounded
to two significant figures
equals 7.2 times 10 to
the negative ninth newtons.
To find the direction
the net force points,
we can use the fact that
the tangent of the angle
the force vector makes
with the positive x axis
equals the sum of the
y components of the forces
acting on q3,
divided by the sum of the
x components of the forces
acting on q3.
If we take the arctangent
of both sides
and then plug in the values
for F sub y and F sub x,
we find that theta,
the angle between the net force
on q3 and the positive x axis
is 84 degrees.
That's a closer look
at advanced applications
of Coulomb's law.
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