1. Electrostatics Superposition: Electric Field

2. 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.

3. Example 1 − + q1 E1 q2 E2 For the following charge configuration
a. Determine the magnitude and direction of the electric field.
0.2 μC P − +
0.4 μC
2 m q1 E1 q2 E2
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.

4. Example 1
For the following charge configuration
a. Determine the magnitude and direction of the electric field.
0.2 μC P − +
0.4 μC
2 m q1 E1
900 q2 E2
450
Now solve for E1 and E2 .
Now that we have the electric field vectors at P , we no longer need the original information given in the diagram.

5. Example 1 E1 E2 900 450 900 450 For the following charge configuration
a. 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
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

6. Example 1 − FE E = 1350 −0.3 μC 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
The new charge is negative. Negative charges move opposite the field (they are the wrong way particle). The direction of force is −x

7. Example 2 − + Find the point where the electric field is zero. 0.1 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.

8. Example 2
Find the point where the electric field is zero. +
0.1 μC −
0.3 μC L S L S L S
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
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)

9. Example 2 q2 q1 − + E2 E1 1 m r1 = x r2 = x + 1
Find the point where the electric field is zero.
0.1 μC
0.3 μC E1 E2 q2 q1 − +
r1 = x
1 m
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

10. Example 2 + − q2 q1 E2 E1 1 m r1 = x r2 = x + 1
Find the point where the electric field is zero. +
0.1 μC −
0.3 μC E1 E2 q2 q1
1 m
r2 = x + 1
r1 = x
To the left of q1

11. Example 3
In the following charge configuration the marks on the axis are 1 m marks. q2 q1 −
a. Determine the magnitude and direction of the electric field at the origin.
0.2 μC
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
3. Label the vectors as E1 and E2 .
4. Solve for each electric field vector.
450
900 E1 E2 +
0.4 μC
I’ll add the values to the diagram, so we can move to the next slide and have everything we nee in the diagram.

12. Example 3
In the following charge configuration the marks on the axis are 1 m marks.
0.2 μC
0.4 μC E1 E2 q2 q1 − +
450
900
a. Determine the magnitude and direction of the electric field at the origin.
5. Add the vectors to find mag. and dir.
When vectors are 90o apart they are added using Pythagorean Theorem, and the direction is found using inverse tan.
Tail to Tail using parallelogram method
Tip to Tail method
900
450 E E
900
450

13. 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. FE +
b. A 0.5 μC charge is inserted at point P . Determine the magnitude and direction of force on this charge.
The new charge appears at point P .
The electric field at P puts a force on the new charge.
Direction of force has its own rules.
Positive particles follow the field.

14. 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.
q1 is positive: field is away from q1
q2 is negative: field is toward q2
3. Label the vectors as E1 and E2 .
4. Solve for each electric field vector.
900 E1 E2 P + −
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.

15. 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.
E = 900 N/C, +x
900
60o
900
900
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.

16. Example 5 − + + − q1 q2 E1 E4 P E3 E2 q4 q3
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.
q1 is negative: field is toward q1
q2 is positive: field is away from q2
q3 is negative: field is toward q3
q4 is positive: field is away from q4
3. Each vector is labeled, to match a charge
4. Solve for each electric field vector.
0.1 μC
0.1 μC q2 q1 q3 q4 − + E1 E4 P E2 E3 + −
0.1 μC
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.

17. Example 6 + − r + − q2 q1 q3 q4 1800 E4 E2 P E1 E3
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.
q1 is positive: field is away from q1
q2 is negative: field is toward q2
q3 is negative: field is toward q3
q4 is positive: field is away from q4
3. Each vector is labeled, to match a charge
4. Solve for each electric field vector.
0.1 μC
0.1 μC q2 q1 q3 q4 + −
1800 E4 E2 r P E1 E3 + −
0.1 μC
0.1 μC
r is the distance from each charge to point P . Solving it involves geometry & a 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

18. 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

19. 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

20. 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
= 3600

21. 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
3600
90o
45o
= 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,
then the answer (including direction) is easy.
Or use Pythagorean theorem
E = 5091 N/C, +x

22. 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 .
The shape is a square.
Note: the length of each side is given.
There is one triangle here.
r is a minor side. Minor sides are
There is another triangle here.
The diagonal line is the hypotenuse and in a the hypotenuse is However, r is only half of the diagonal.
sides that are 1 m long r P
You can also find the diagonal with Pythagorean theorem. Just remember that r is half the diagonal.