26: PSS 26.1: The Electric Field of Multiple Point Charges

INTRO:
Three positively charged particles, with charges q₁=q, q₂=2q, and q₃=q (where q>0), are located at the corners of a square with sides of length d. The charge q₂ is located diagonally from the remaining (empty) corner.

Find the magnitude of the resultant electric field Enet in the empty corner of the square.

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PART A:
Below is an incomplete pictorial representation of the situation described in this problem. Complete the sketch by drawing the electric field due to each charge at point P. Make sure that all your vectors have the correct orientation.
The orientation of your vectors will be graded. The length of your vectors will not be graded.

SOLUTION:
Very obvious. The direction of the electric field is the same as the connecting line between the charge and point P

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PART B:
Now, consider the resultant electric field E_net at P. With reference to the coordinate system shown in the previous part, which component of E_net, if any, is zero in this problem?

• only the x component
• only the y component
• both the x and y components
• neither the x nor the y component

SOLUTION:
none of the components is zero, so the last option is correct
neither the x nor the y component

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PART C:

Determine the magnitude E_net of the net electric field at point P. Use K for the electrostatic constant.
Express your answer in terms of q, d, and K.

SOLUTION:

Using equation (26.1) E = kq/r², all components of all electric fields can be determined. 

the electric field due to q₃ at point P is just an x-component of a field...

the distance in consideration is just d, so
E₃ = E_3x =kq/d²

the electric field due to q₂ at point P is both an x-component and a y-component of a field...

since the setup is that of a square, the magnitude of the x-component is equal to the magnitude of the y-component. 
E_2x = E₂cos(45) = E₂cos(45) = E₂⋅1/√2 

E_2x = kq√2/(2d²)

the y-component is the same but in the y-direction.

E_netx = E_3x + E_2x 

= kq/d² + (√2/2)kq/d²
⇒E_netx =(1+√2/2)kq/d²

the y-magnitude of this net electric field is the same as the x-magnitude, but in the y-direction. 
total mag = √(x+y)

& if |x| = |y| ... then total mag = x⋅√2 = y⋅√2

→ E_net = √2⋅(1+√2/2)kq/d²

⇒ E_net =(1+√2)kq/d²

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PART D:

Intuitively, which of the following would happen to E_net if d became very large?

1. A) E_net should reduce to the field of a point charge of magnitude q.
2. B) E_net should reduce to the field of a point charge of magnitude 4q.
3. C) The larger d becomes, the smaller the magnitude of E_net will be.
4. D) The larger d becomes, the greater the magnitude of E_net will be.

Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if you think that A and D are correct, enter AD.

SOLUTION:

C) The larger d becomes, the smaller the magnitude of E_net will be

NOTE:

Looking at your answer derived in Part C, you should see that the magnitude Enet will decrease as d gets larger, just as you would expect. Your results do make sense! It is interesting to notice that the field will not reduce to that of a point charge (of magnitude 4q) unless the point P is moved farther away from the three charges while they themselves remain at their original positions.

26: Charged Ring

INTRO:
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius a and positive charge q distributed evenly along its circumference. (Figure 1)

Figure 1

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PART A:
What is the direction of the electric field at any point on the z axis?

• parallel to the x axis
• parallel to the y axis
• parallel to the z axis
• in a circle parallel to the xy plane

SOLUTION:
Parallel to the z-axis

NOTES:
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring. In what direction is the net field? What if you did this for every pair of points on opposite sides of the ring?

Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy about its axis. As a result of the rotation, the component of the electric field in the xy plane will rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is rotated looks just like the one that isn't. However, they have the component of the electric field in the xy plane pointing in different directions! This apparent contradiction can be resolved if this component of the field has a particular value. What is this value?
Does a similar argument hold for the z component of the field?
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PART B:
What is the magnitude of the electric field along the positive z axis?
Use k in your answer, where k=1/(4πϵ₀).

SOLUTION:
Use Coulomb's law, F=k⋅q₁⋅q₂/r², to find the electric field (the Coulomb force per unit charge) due to a point charge. 
Given the force, the electric field at q₂ due to q₁ is E=F/q²=k⋅q₁/r².

Next, use Coulomb's law to find the contribution dE to the electric field at the point (0,0,z) from a piece of charge dq on the ring at a distance r away. Then, you can integrate over the ring to find the value of E. Consider an infinitesimal piece of the ring with charge dq. Use Coulomb's law to write the magnitude of the infinitesimal dE at a point on the positive z axis due to the charge dq shown in the figure.

∴ dE = kdq/(z² + a²)

By symmetry, the net field must point along the z axis, away from the ring, because the horizontal component of each contribution of magnitude dE is exactly canceled by the horizontal component of a similar contribution of magnitude dE from the other side of the ring. Therefore, all we care about is the z component of each such contribution. 
The component dE_z of the electric field caused by the charge on an infinitesimally small portion of the ring dq in the z direction is... 
dE_z = dE_z/(z² + a²)^½

Next, integrate around the ring. 
By combining the previous two results, you will have an expression for dE_z, 
the vertical component of the field due to the infinitesimal charge dq. 

The total field is
E =E_zk^=k^∮_ringdE_z.


If you are not comfortable integrating dq over the ring, change to a spatial variable. Since the total charge q is distributed evenly about the ring, convince yourself that

∮_ringdq=∫₀^2πq/(2π) dθ.

->    E(z) = kzq/(z² + a²)^3/2

NOTE: 
Notice that this expression is valid for both positive and negative charges as well as for points located on the positive and negative z axis. If the charge is positive, the electric field should point outward. For points on the positive z axis, the field points in the positive z direction, which is outward from the origin. For points on the negative z axis, the field points in the negative z direction, which is also outward from the origin. If the charge is negative, the electric field should point toward the origin. For points on the positive z axis, the negative sign from the charge causes the electric field to point in the negative z direction, which points toward the origin. For points on the negative z axis, the negative sign from the z coordinate and the negative sign from the charge cancel, and the field points in the positive z direction, which also points toward the origin. Therefore, even though we obtained the above result for postive q and z, the algebraic expression is valid for any signs of the parameters. As a check, it is good to see that if |z| is much greater than a the magnitude of E(z) is approximately kq/(z²), independent of the size of the ring: The field due to the ring is almost the same as that due to a point charge q at the origin.

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PART C:

Imagine a small metal ball of mass m and negative charge −q₀. The ball is released from rest at the point (0,0,d) and constrained to move along the z axis, with no damping. If 0<d≪a, what will be the ball's subsequent trajectory?

• repelled from the origin
• attracted toward the origin and coming to rest
• oscillating along the z axis between z=d and z=−d
• circling around the z axis at z=d

SOLUTION:

oscillating along the z axis between z=d and z=−d

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PART D:

The ball will oscillate along the z axis between z=d and z=−d in simple harmonic motion. What will be the angular frequency ω of these oscillations? Use the approximation d≪a to simplify your calculation; that is, assume that d²+a²≈a².
Express your answer in terms of given charges, dimensions, and constants.

SOLUTION:

Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the system states that
F_x=m⋅d²x/dt²=−k′x, leading to oscillation at a frequency of ω=√k′/m

(here, the prime on the symbol representing the spring constant is to distinguish it from k=1/(4πϵ₀). The solution to this differential equation is a sinusoidal function of time with angular frequency ω. Write an analogous equation for the ball near the charged ring in order to find the ω term.

Next find the force on the charge
What is F_z, the z component of the force on the ball on the ball at the point (0,0,d)? Use the approximation d²+a²≈a².

F_z = (-k⋅q⋅q₀⋅d)/a³

^-->     ω = [(k⋅q⋅q₀)/(a³⋅m)]^½

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26: Problem 26.4

INTRO:
What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

Figure 1

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PART A:
Specify the strength of the electric field. Let r = 5.5 cm .
Express your answer using two significant figures.

SOLUTION:
<< explanation to be added >>
E = 6300 N/C

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PART B:
Specify the direction.
Express your answer using two significant figures.

SOLUTION:
<< explanation to be added >>
θ = 90° below horizontal

Chapter 26: Overview

CHAPTER 25: THE ELECTRIC FIELD

Coulomb’s law:

F = K⋅|q₁|⋅|q₂| / r²

Electric field of a point charge
E = K ⋅ q/r² ⋅ (r^)

Useful:

Electric constant
ε₀ = 8.854×10^-12 F/m

SI Unit conversions
1 N = 1 kg⋅m⋅s^-2
1 C = A⋅s
1 F = A²⋅s⁴⋅kg^-1⋅m^-2

26: PhET Tutorial: Charges and Electric Fields

Part A: 
The electric field produced by the positive charge

a) is directed radially toward the charge at all locations near the charge.
b) wraps circularly around the positive charge.
c) is directed radially away from the charge at all locations near the charge.

SOLUTION:
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge.

Part B:
For these four locations, the magnitude of the electric field is

a) greatest to the left of the charge.
b) the same.
c) greatest below the charge.
d) greatest to the right of the charge.
e) greatest above the charge.

SOLUTION:
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric.

Part C:
The magnitude of the electric field 1 m away from the positive charge is 
a) one-half
b) equal to
c) two times
d) four times
e) one-quarter
the magnitude of the electric field 2 m away.

SOLUTION:
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (E∝1/r², where r is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulomb'€™s law, which states that the magnitude of the force between two charged particles is F=kQ₁Q₂/r².

Part D:
If the field strength is E = 9 V/m a distance of 1 m from the charge, what is the field strength E a distance of 3 m from the charge?

SOLUTION:
E∝1/r², so E₂ = 9/3² = 1 V/m

Part E:
Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge?

a) The electric field changes direction (now points radially inward), but the electric field strength does not change.
b) Nothing changes; the electric field remains directed radially outward, and the electric field strength doesn'™t change.
c) The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.

SOLUTION:
The electric field is now directed toward the negative charge, but the field strength doesn'€™t change. The electric field of a point charge is given by E⃗ =(kQ/r²)r^. Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same.

Part F:
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?

a) The electric field is nonzero everywhere on the screen.
b) The electric field is roughly zero near the midpoint of the two charges.
c) The electric field is zero at any location along a vertical line going through the point directly between the two charges.

SOLUTION: 
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero.

Part G:
Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/m. What is the magnitude of the total electric field due to both charges at this location?

a) 36 V/m
b) 25 V/m
c) zero

SOLUTION:
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.

Part H:
The electric field at the midpoint is

a) directed to the left
b) zero
c) directed to the right

SOLUTION:
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right.

Part I:
Measure the strength of the electric field 0.5 m directly above the midpoint as well as 1 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r³! So the field 1 m above the midpoint is roughly eight times weaker than at 0.5 m above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge.

Part J:
Measure the strength of the electric field 1 m directly above the middle as well as 2 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r. So the field 1 m above the midpoint is roughly half the strength at 0.5 m. This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.

26: Electric Field due to Multiple Point Charges

INTRO:
Two point charges are placed on the x axis. The first charge, q₁ = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q₂ = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Part A: 
Calculate the electric field at point A, located at coordinates (0 m, 12.0m).
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

SOLUTION:
Find the contributions to the electric field at point A separately for q₁ and q₂, then add them together using vector addition to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A.

Triangle 1 ⇒ ΔOAq₁
a = 12 m; b = 16 m
∴c = 20 m
Triangle 2 ⇒ ΔOAq₂
a = 12 m; b = 9 m
∴c = 15 m

Now find the contributions to the electric field at point A
E_A1 = K⋅q₁/r² = K⋅(8.00 nC)/(20 m)² = 0.1798 N/C & points in direction <-16, 12>
E_A2 = K⋅q₂/r² = K⋅(6.00 nC)/(15 m)² = 0.2397 N/C & points in direction <9, 12>

Now calculate the x and y components of both electric fields 
to get the angle for E_A1, θ = tan^-1(12/16) = 36.87°
E_A1x = E_A1 cos(θ) = 0.1798 N/C⋅0.8 = 0.14384 N/C
E_A1y = E_A1 sin(θ) = 0.1798 N/C⋅0.6 = 0.10788 N/C
E_A1 = <-0.14384, 0.10788> N/C
to get the angle for E_A2, θ = tan^-1(12/9) = 53.13°
E_A2x = E_A2 cos(θ) = 0.2397 N/C⋅0.6 = 0.14382 N/C
E_A2y = E_A2 sin(θ) = 0.2397 N/C⋅0.8 = 0.19176 N/C 
E_A2 = <0.14382, 0.19176> N/C

SO, now that you have the vector components of the two electric fields, you are able to use vector addition to determine the net electric field at point A
E_NET = E_A1 + E_A2 = <-0.144, 0.108> N/C + <0.144 0.192> N/C = <0, 0.300> N/C

Part B:
An unknown additional charge q₃ is now placed at point B, located at coordinates (0 m, 15.0m).

Find the magnitude and sign of q₃ needed to make the total electric field at point A equal to zero.
Express your answer in nanocoulombs to three significant figures.

SOLUTION:
luckily q₃ is in line with point A so we don't need to worry about x and y components as with the previous part. 
givens:
r_AB = 3 m
E_AB = <0, 0.300> N/C

E_AB = K q₃/r_AB²
q₃ = E_AB⋅r_AB² / K
→0.300 N/C ⋅ (3 m)² / [4πε₀]
q₃ = 3×10^-10 C = 0.300 nC