Download Electric Charge

Superposition: Electric Field
Superposition
Finding the electric field when there is more than one charge
surrounding point P.
Solve one charge at a time.
Find the magnitude (value) and direction of each charge.
This will give you the electric field vectors for every charge in
the diagram.
Then use vector addition to add all the electric field vectors to
find the total electric field.
Example 1
a.
For the following charge configuration
Determine the magnitude and direction of the electric field.
0.4 μC
P
q1 +
0.2 μC
E1
− q2
E2
2m
2m
I recommend that you draw the field vectors created by each charge before
you start calculating.
Visualize an imaginary test charge at point P .
Ask yourself which direction the test charge will move due to each charge in
the diagram.
The positive charge (let’s call it q1) will repel the test charge to the right.
E1 (caused by q1) is to the right.
The negative charge (let’s call it q2) will attract the test charge to the right.
E2 (caused by q2) is to the right.
Example 1
a.
For the following charge configuration
Determine the magnitude and direction of the electric field.
0.4 μC
P
q1 +
0.2 μC
E1
900
− q2
E2
450
2m
2m
Now solve for E1 and E2 .
E1 = k
q1
r12
(
= ( 9 ´ 10 )
9
0.4 ´ 10-6
()
( 2)
2
) = 900 N C
)
-6
0.2
´
10
q2
E2 = k 2 = 9 ´ 10 9
= 450 N C
2
r2
( 2)
Now that we have the electric field vectors at P , we no
need the original information given in the diagram.
(
Example 1
a.
For the following charge configuration
Determine the magnitude and direction of the electric field.
P
E1
900
E2
450
This is now a vector addition problem (see summer work)
This vectors have the same direction, so they can be added very easily.
900 + 450 = 1350 N C
900
450
We still need the direction of the electric field.
This time we are lucky, there is no fancy trig involved. Just look at the arrows
+x
Example 1
For the following charge configuration
b. A −0.3 μC charge is inserted at point P .
Determine the magnitude and direction of the force on this charge.
P
FE
−
E = 1350
−0.3 μC
The electric field solved in part a (due to the original charges) pulls on
the new charge
(
FE = qE = 0.3 ´ 10
-6
) (1350) = 4.05 ´ 10
-4
N
The new charge is negative. Negative charges move opposite the field (they
are the wrong way particle). The direction of force is
−x
Example 2
Find the point where the electric field is zero.
0.1 μC
0.3 μC
−
+
Before solving mathematically we need to know approximately where the
point will be. There are three possible locations. First draw the vectors for
the electric field, due to each charge, at all three locations. (Remember: you
can place a positive test charge at all three points and ask which way it moves
to determine field direction)
The negative charge (black) will create a field that is toward itself at all three
points. ( + test chg would move toward the negative charge)
The positive charge (red) will create a field directed away from itself at all
three points. ( + test chg would move away from the positive charge)
The middle point can now be ruled out, since those two arrows cannot cancel.
Let’s take a closer look at the two remaining points.
Example 2
L
S
S
L
Find the point where the electric field is zero.
0.1 μC
0.3 μC
−
+
S
S
L
L
Now look at the equation to find the correct location.
Charge q and electric field E are directly related.
0.1 μC charge is small: Field is small (S) field
q
E=k 2
r
0.3 μC charge is large: Field is large (L) field.
Distance r and electric field E are inversely related.
0.1 μC charge is close to the left point: Field is large (L)
0.1 μC charge is far from the right point: Field is small (S)
0.3 μC charge is far to the left point: Field is small (S)
0.3 μC charge is close to the right point: Field is large (L)
It has to be the left point (L & S combine to produce cancelling medium fields)
The right is impossible (XL cannot be cancelled by XS)
Example 2
Find the point where the electric field is zero.
0.1 μC
E2
E1
0.3 μC
− q1
+ q2
1m
r1 = x
r2 = x + 1
Let’s define the 0.1 μC charge as q1 producing E1 , and the 0.3 μC charge
as q2 producing E2 .
In the problem on the worksheet it states that the charge are 1 m apart.
We also need to identify the distance from each charge to point P .
r1 is unknown: set it equal to x .
This makes r2 = (x + 1 m)
The next step is the math: E1 = E2
Example 2
Find the point where the electric field is zero.
0.1 μC
E2
E1
0.3 μC
− q1
1m
r1 = x
r2 = x + 1
E1 = E2
q1
q2
k 2 =k 2
r1
r2
(0.1 ´ 10 ) = (0.3 ´ 10 )
-6
-6
x2
(
0.1´ 10 -6
x2
( x + 1)
-6
0.3´
10
(
)
2
)=
+ q2
( x + 1) 2
3.16 ´ 10 -4 5.48 ´ 10 -4
=
( x + 1)
x
3.16 ´ 10-4 ( x + 1) = 5.48 ´ 10-4 x
3.16 ´ 10-4 x + 3.16 ´ 10-4 = 5.48 ´ 10-4 x
3.16 ´ 10-4 = 2.32 ´ 10-4 x
x = 1.36 m
To the left of q1
Example 3
a.
In the following charge configuration the marks on the axis
are 1 m marks.
0.2 μC − q1
Determine the magnitude and direction
of the electric field at the origin.
1. Identify charges as q1 and q2 .
2. Draw the field vectors of each charge.
q1 is negative: field is toward q1
q2 is positive: field is away from q2
q2
+
0.4 μC
E1
450
E2
900
3. Label the vectors as E1 and E2 .
4. Solve for each electric field vector.
E=k
q
r2
E1
-6
0.2
´
10
(
)
9
= ( 9 ´ 10 )
= 450
E2
-6
0.4
´
10
(
)
9
= ( 9 ´ 10 )
= 900
( 2)
2
( 2) 2
I’ll add the values to the
diagram, so we can move to the
next slide and have everything
we nee in the diagram.
Example 3
In the following charge configuration the marks on the axis
are 1 m marks.
0.2 μC − q1
a. Determine the magnitude and direction
of the electric field at the origin.
5. Add the vectors to find mag. and dir.
q2
E1
450
When vectors are 90o apart they are added
+
using Pythagorean Theorem, and the
E2
900
direction is found using inverse tan.
0.4 μC
Tail to Tail using
parallelogram method
450
Tip to Tail
method
E
E
900
900
Etotal =
450
( 450) + ( 900) = 1006
2
2
q = tan
Etotal = 1006 N C, 26.6o
-1
450
= 26.6 o
900
Example 3
In the following charge configuration the marks on the axis
are 1 m marks.
The result of all the previous work is the
electric field at point P . Now we no longer
need any of the previous information. Once
we have the electric field it makes the rest
of the problem easy.
Etotal = 1006 N C, 26.6 o
b. A 0.5 μC charge is inserted at point P .
Determine the magnitude and direction
of force on this charge.
+
FE
The new charge appears at point P .
The electric field at P puts a force on the new charge.
FE = qE
(
)
FE = 0.5´ 10 -6 (1006)
Direction of force has its own rules.
Positive particles follow the field.
FE = 5.03 ´ 10-4 N, 26.6o
Example 4
Determine the electric field at point P . The arrangement
forms an equilateral triangle, with sides that are 1 m long.
1. Identify charges as q1 and q2 .
2. Draw the field vectors of each charge.
E1
900
q1 is positive: field is away from q1
P
q2 is negative: field is toward q2
3. Label the vectors as E1 and E2 .
4. Solve for each electric field vector.
E2
900
q
E=k 2
r
-6
0.1´
10
(
)
9
E = ( 9 ´ 10 )
= 900
(1)
2
+
−
0.1 μC
0.1 μC
q1
q2
The minus sign on q2 is used in step 2 above to find vector direction. It is not used
in the calculation of magnitude (numerical value).
q1 and q2 have the magnitude of charge and are the same distance (1 m) from P .
As a result, there is no reason to repeat the calculation for the second charge, as it
will be identical to the first.
Example 4
Determine the electric field at point P . The arrangement
forms an equilateral triangle, with sides that are 1 m long.
5. Add the vectors to find the total electric field.
These vectors are tail to tail, making the
parallelogram method ideal.
In this method the resultant is drawn from the tails to
the intersection of the parallelogram.
Find the magnitude and direction of the green arrow
and you have your answer.
If you look at all the angles (see geometry of
equilateral triangles),
then the answer (including direction) is easy.
60o
60o
900
900
60o
60o
60o
60o
60o
900
60o
900
60o
E = 900 N/C, +x
When you draw short cuts reveal themselves.
If you don’t see the short cuts, then you have to do it the hard way.
Add x-components, add y-component, Pythagorean theorem x & y , and tan−1.
Example 5
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
1. Identify charges as q1 and q2 .
2. Draw the field vectors of each charge.
0.1 μC q
−
q2
1
0.1 μC
+
q1 is negative: field is toward q1
q2 is positive: field is away from q2
q3 is negative: field is toward q3
E1
q4 is positive: field is away from q4
P
3. Each vector is labeled, to match a charge
4. Solve for each electric field vector.
E=0
E4
E2
+
0.1 μC
q4
E3
q3
−
0.1 μC
I recommend drawing the individual electric field vectors first. If there is a short
cut it will be revealed before you do needless work. The charges all have the same
magnitude and are the same distance from P . There is no point in solving in
wasting time with the equation. We can see that the arrows cancel. The diagram is
enough work to get grade points.
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
1. Identify charges as q1 and q2 .
2. Draw the field vectors of each charge.
0.1 μC q
q2
1
+
0.1 μC
−
q1 is positive: field is away from q1
q2 is negative: field is toward q2
1800
E4
q3 is negative: field is toward q3
r
q4 is positive: field is away from q4
P
3. Each vector is labeled, to match a charge
4. Solve for each electric field vector.
q
E=k 2
r
-6
0.1
´
10
(
)
9
E = ( 9 ´ 10 )
= 1800
æ 2ö
ç 2 ÷
è
ø
2
E
1800
2
E1800
1
E3
1800
+
0.1 μC
q4
q3
−
0.1 μC
r is the distance from each charge to point P . Solving it
involves geometry & a 45-45-90 triangle. The r from q1
is shown. All charges are the same distance r from P.
There is no need to do the calculation 4 times. The charges and distance are the same
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
5. Add vectors to find the total electric field.
The easiest vectors to add up are vectors that
point the same way. You just add them.
1800
1800
1800
1800
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
5. Add vectors to find the total electric field.
The easiest vectors to add up are vectors that
point the same way. You just add them.
1800
1800
1800
1800
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
5. Add vectors to find the total electric field.
The easiest vectors to add up are vectors that
point the same way. You just add them.
3600
1800 + 1800 = 3600
1800 + 1800 = 3600
3600
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
5. Add vectors to find the total electric field.
The easiest vectors to add up are vectors that
point the same way. You just add them.
1800 + 1800 = 3600
1800 + 1800 = 3600
These vectors are tail to tail, making
the parallelogram method ideal.
The resultant is drawn from the tails to the
intersection of the parallelogram. Solve for
the green vector.
If you look at all the angles,
90o
3600
3600
45o
45o
45o
45o
3600
3600
90o
then the answer (including direction) is easy.
E = hyp = 2 ´ side = 2 ´ 3600 = 5091
E = 5091 N/C, +x
Or use Pythagorean theorem
Example 6
Determine the electric field at point P at the center of the
arrangement, a square with sides that are 1 m long.
For those who can’t find r .
1.The shape is a square.
2.Note: the length of each side is given.
3.There is one 45-45-90 triangle here.
• r is a minor side. Minor sides are
sides that are 1 m long
r
P
2
2
2
r=
´ Hyp =
´1=
2
2
2
4.There is another 45-45-90 triangle here.
• The diagonal line is the hypotenuse and
in a 45-45-90 the hypotenuse is
diagonal = 2 ´ Side
However, r is only half of the diagonal.
diagonal
2 ´ Side
2 ´1
2
r=
=
=
=
2
2
2
2
5.
You can also find the
diagonal with
Pythagorean theorem.
Just remember that r
is half the diagonal.