Near the surface of the earth, we agreed to call the force of gravity of constant.

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Roderick Lamb
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Electric Fields 1. A field 2. Field lines 3. The Electric Field 4. Field from a dipole 5. Line charge 6.

This meant that, to a reasonable approximation, the force between an object, like a rock, and the earth was independent of their separation.

A field is a physical quantity that has a value at every point in space and time. 2. A field is a mathematical function of space and time. 3.

1 Electric Fields 1. A field 2. Field lines 3. The Electric Field 4. Field from a dipole 5. Line charge 6. Other configurations Near the surface of the earth, we agreed to call the force of gravity of constant. This meant that, to a reasonable approximation, the force between an object, like a rock, and the earth was independent of their separation. Later, we saw that the force is actually dependent on the distance between the two object. (As well as their masses) F = G m 1 m 2 r 2 A field some definitions 1. A field is a physical quantity that has a value at every point in space and time. 2. A field is a mathematical function of space and time. 3. A field is the alteration of space. 4. A distribution in a region of space of the strength and direction of a force Scalar Field A scalar field has a value at every point, but no direction associated with it. The gravitational potential also varied with position. Fig. 2 updated on J. Hedberg 2018 Page 1

Scalar field A temperature field might tell us what the temperature is as a

At every point in space there is a value for the temperature. T(x,y, z) Fig.

earth around the country: Wind Field velocity of blood through arteries 3 2 1

The direction of the vector indicates the direction of the field.

2 Scalar field A temperature field might tell us what the temperature is as a function of position in the room. At every point in space there is a value for the temperature. T(x,y, z) Fig. 3 Vector fields wind around a wing river currents gravitational force Wind earth around the country: Wind Field velocity of blood through arteries The size of the vector indicates the strength of the field at that point. The direction of the vector indicates the direction of the field. Here, we ve picked equally spaced points for the graph. F = 1 r 2 r^ Fig. 5 updated on J. Hedberg 2018 Page 2

3 Field lines Or, we can draw field lines. Each of these lines are tangent to the direction of the field. The strength of the field in a given area is seen by looking at the density of field lines. earth Another field This says, the force in the x direction is equal to the x coordinate, and the force in the y direction is always zero. F x = x = 0 F y You can see the vectors getting longer as we move further away from the y-axis (x is getting bigger) Fig. 6 updated on J. Hedberg 2018 Page 3

4 Quick Question 1 Which field diagram represents the following function: F x = x = y F y Quick Question 2 What could this field be defined by? 1. F x = 0 ; = 1 F y 2. F x = x ; = y F y 3. F x = 1 ; = +y F y 4. F x = 1/x ; = 1/y F y Force from a single charge Let's consider Coulomb's Law q F 1 on 2 = k 1 q 2 r 2 updated on J. Hedberg 2018 Page 4

5 +y -x +x -y The Electric Field The field is defined as the force at a point (x,y, z) on a charge q 0. E (x, y, z) = F on q 0 q 0 (x, y,z) q 0 can be thought of a test charge or a probe charge. Essentially we will use it to examine the field strength at various points in space. The SI unit of an electric field are N/C The field from a point charge E q 0 = F We know the force is directed away from the source charge Q and has the magnitude: Q q 0 F Q on q0 = k From this, we can say that the field strength at the point is: r 2 F E = e = q 0 kq r 2 So, we can see the field strength does not depend on the test charge. q 0 updated on J. Hedberg 2018 Page 5

6 Quick Question 3 source charge +1e A) +1e test charge Here are four source and test charge arrangements. Which has the highest field strength at the location of the test charge? B) C) +2e +3e +5e +3e D) +4e +1e Field Directions To determine the direction of the field, we will say: 1) if the test charge is positive, the field points in the same direction as the force. 2) if the test charge is negative, the field points in the opposite direction as the force. charge would move if allowed to. And so, we can say, in general, that the field lines point in the direction that a positive Two point charges These are just representations of a few field vectors. (The field is actually defined at every point in space.) updated on J. Hedberg 2018 Page 6

7 Field vectors and field lines provide two means of visualizing a field in space About the representations field vectors field lines 1. The field lines are tangent to the field vectors 2. The field lines are denser in regions where the fields are stronger 3. The field lines start from positive, and end on negative. updated on J. Hedberg 2018 Page 7

8 4. The field lines land perpendicular to a charged objects surface. Usually, there s more than one charge... It might be some arrangement of charges. But, the same prescription applies. Just find the force on a test charge at every point in space, then the field is known. Field from a dipole +y Two opposite charges near each other create a dipole. d -q Fields from each charge can be figure out, and added together given the resultant field. E total = E + + E +q +x Example Problem #1:.. at point P along the x axis. -q +q P d updated on J. Hedberg 2018 Page 8

9 Quick Question 4 +y -q P +x +q Which direction is the electric field at point P? 1. +x 2. -x 3. +y 4. -y 5. +z 6. -z updated on J. Hedberg 2018 Page 9

10 A dipole field would look like this. Dipole Moment The field from a dipole along the dipole axis was found to be: 1 qd E dipole = 2πϵ 0 z 3 We can call the quantity qd the dipole moment. This is essentially a vector pointing from the negative charge to the positive charge with a magnitude qd. -q +q d updated on J. Hedberg 2018 Page 10

11 1 E dipole = 2πϵ 0 p z 3 The dipole moment p, can then tell us everything about the dipole: its strength and orientation. Quick Question 5 +q -q -q +q Four charges are located on the corners of a square as shown in the drawing. What is the direction of the net electric field at the center of the square? 1. toward the upper left corner of the square 2. toward the middle of the right side of the square 3. toward the middle of the bottom side of the square 4. toward the lower right corner of the square 5. There is no direction. The electric field at the center is zero N/C. updated on J. Hedberg 2018 Page 11

12 Quick Question 6 +2q -q -2q +q Four charges are located on the corners of a square as shown in the drawing. What is the direction of the net electric field at the center of the square? 1. toward the upper left corner of the square 2. toward the middle of the right side of the square 3. toward the middle of the bottom side of the square 4. toward the lower right corner of the square 5. There is no direction. The electric field at the center is zero N/C. A charge and field simulator Quick Question 7 Could you solve this indefinite integral? dx ( a 2 + x 2 ) (3/2) 1. Yes, with my eyes closed 2. Yes, but it would take like 15 minutes 3. Maybe 4. Probably not without some help 5. Not a chance Line charge Many situation will involve a continuous distribution of charges, as opposed to discrete bits of charge. For example, a line of charge. updated on J. Hedberg 2018 Page 12

13 P Find the field at a point P, which is above the center of the line of charge. y +L/2 L = length x -L/2 If the line charge extends to infinity, updated on J. Hedberg 2018 Page 13

14 y infty = length x E infinite line 2 λ = k r Quick Question 8 Consider a line of charge of length L, that has a linear charge density λ and is lying along the x axis beginning at x = d. Which one of the following expressions allows one to calculate the electric field at the origin? E = E = E = E = E = λ L 4πϵ 0 d dx x 2 λ L 4πϵ 0 λ d+l 4πϵ 0 d 0 x2 λ L 4πϵ 0 0 dx x λ d+l 4πϵ 0 0 dx dx x 2 dx x Example Problem #2: There is Q charge on this uniform linear charge. Find the electric field at the origin. d L x=0 updated on J. Hedberg 2018 Page 14

15 Other configurations Example Problem #3: An insulating partial ring has charge +Q deposited on it. Find the electric field at point O. 60º r x Example Problem #4: Find the field above the center point of this ring of charge. The ring has a radius R and a linear charge density λ on it. Example Problem #5: Integrate a ring to find the area of a circle. Example Problem #6: Find the field above the center of this charged disc with surface charge density σ and radius R. updated on J. Hedberg 2018 Page 15

16 Charges in motion 'member F = ma? Newton's laws still hold for charges. (for the most part) Now, we ll have to calculate Forces, based on fields or charge arrangements, in order to determine accelerations of charged particles. Here's a side view of two charged plates. Inside the plates, the field can be considered uniform (if the plates have the right geometry) updated on J. Hedberg 2018 Page 16

17 Quick Question 9 An electron enters a region of uniform electric field as shown. Which of the following trajectories would it follow? A B C E D Example Problem #7: Calculate the acceleration of a charged particle in a uniform field of strength 30 N/C. The particle has a mass of 0.15 g and has a charge of C. (Ignore gravity and other forces, for now!) Let's put a dipole in a uniform electric field. -q +q +q -q Since the dipole has no net charge, there will be no net force on the dipole. However, there may still be a torque. This torque can be calculated by using the dipole moment, p. The magnitude of this torque is then: τ = p E updated on J. Hedberg 2018 Page 17

18 τ = pe sinθ ( θ is the angle between the dipole and the electric field lines) However, we should note that this is really a restoring force, in that the torque causes the dipole to align with the field no matter which way it's pointing. Let's add a negative sign to indicate this. τ = pe sinθ Dipole in a field A dipole in an electric field will experience a torque based on the relation: τ = p E = pe sinθ Your physics intuition should be reminded of a pendulum. Example Problem #8: What would be the oscillation frequency for a dipole with a dipole moment p and moment of inertia I in an electric field E? Potential Energy of a Dipole in an Electric Field θ θ U = W = τdθ = pe sinθdθ Which, after doing the requisite integration becomes: U dipole = pe cos θ = p E Quick Question 10 Which of these dipoles has the most potential energy? (E if all have the same) A B C D updated on J. Hedberg 2018 Page 18

21.4 Electric Field and Electric Forces

21.4 Electric Field and Electric Forces How do charged particles interact in empty space? How do they know the presence of each other? What goes on in the space between them? Body A produces an electric