Quiz 1 (lecture 2) * A positive and negative charge with equal magnitude are connected by a rigid rod, and placed near a large negative charge. What is the direction of the net force on the two connected charges? Left – when big charge is on left Positive charge is attracted (left) Negative charge is repelled (right) Positive charge is closer so force to left is larger - + - (A) Left (B) Zero (C) Right

The Electric Field + + q1 q2 Coulomb’s Law of Electro-static Force: Fix a positive charge q1 in place and put second positive charge near it. From Coulomb’s Law, we know that q1 exerts a repulsive electrostatic force on q2. Action at a distance occurs because q1 sets up an electric field in the space surrounding. At any given point P in space, the field has both magnitude and direction. The magnitude depends on the magnitude of q1 and the distance between P and q1. The direction depends on the direction from P to q1 and the electrical sign of q1. Thus, when we place q2 at P, q1 interacts with q2 through the electric field at P. The magnitude and direction of that electric field determine the magnitude and direction of the force acting on q2. How does q1 know of the presence of q2? q2 sets up an electric field in the space surrounding it. At any point the field has both a magnitude and direction.

A Scalar Field 77 82 83 68 55 66 75 80 90 91 71 72 84 73 88 92 64 Scalar Fields: temperature distribution, pressure distribution. These isolated temperatures sample the scalar field T, but T is defined (and can be measured) everywhere (x,y)

A Vector Field 77 82 83 68 55 66 75 80 90 91 71 72 84 73 57 88 92 56 64 It may be more interesting to know which way the wind is blowing and how fast.

The Electric Field The electric field is a vector field: consists of a distribution of vectors, one for each point in the region around a charged object. Electric field at point P + Test charge qo at point P + Can define an E field at some point near a charged object, such as P in Figure. Place a positive test charge qo at P. Measure the electrostatic force F that acts on the test charge. Then, define the electric field E at point P due to the charged object as E=F/qo and the direction of E is the direction of F. Tail of E vector is on P. The field exists independent of the test charge. The field at point P existed before and after the test charge was put there. We assumed that the presence of the test charge does not affect the charge distribution on the charged object and thus does not alter the electric field. Define: Magnitude: Direction of that acts on a positive test charge Direction: SI Unit: N/C

Electric Field Observation: The net Coulomb force on a given charge is always proportional to the strength of that charge. q1 F1 F q0 Using principal of superposition for F. F2 q2 test charge Define the electric field, which is independent of the test charge, q0, and depends only on position in space: q1 and q2 in F are the sources of the electric field, q0 is the test charge

Electric Field due to Multiple Point Charges To find the resultant field from n point charge: so the electric field is, by definition, given by Principle of Superposition!

Electric Field – Summary Electric field is generated by any charged object. Electric field is a vector field and obeys the principle of superposition, i.e., the field of a system of charged objects is equal to the (vector) sum of the field of each individual charged object in the system. The electrostatic force between charged objects is mediated by the electric field.

Electric Field Lines A visualization tool to illustrate the geometry of an electric field. Electric field lines originate from positive charges and terminates at negative charges. The direction of the electric field at any location is tangential to the field line there. The magnitude of the electric field at any location is proportional to the density of the lines there. At any point, the direction of a straight field line or the direction of the tangent to a curved field line fives the direction of E at that point. Field lines are drawn such that the number of lines per unit are, measured in a plane that is perpendicular to the lines, is proportional to the magnitude of E. (where line are close E is large. Where lines are far apart E is small). The density of field lines is an arbitrary choice, used to visualize a picture of the field. Negatively charged uniform sphere. Put qo positive test charge at all points. Get above figure. Spreading lines with distance tells us that the magnitude of E decreases with distancce from the sphere. If a positive sphere was used instead, the lines would point away from the sphere.

Electric field due to a Point Charge • represents various positive test charges. represents various electric field vectors. Note the change in length of with distance r from the positive charge. Direction of F is always away from positive charge (and always toward negative charge). The force acting on a test charge, qo is: The electric field at any point a distance r from the charge,q :

Electric Dipole: opposite signs but equal magnitude Electric Field Lines Electric Dipole: opposite signs but equal magnitude Two Positive Charges: with equal magnitude On opposite charges with unequal magnitudes: Opposite charges. Note that twice the number of flux lines enter (or leave) the charge that is twice as large DEMOS: Plumes (E-9) and wheat + parallel plates DEMO

Electric Field Lines DEMOS: Plumes (E-9) and wheat + parallel plates Opposite Charges: unequal magnitude Far from charges, the field lines are as if they are due to a point charge of +2q-q=+q On opposite charges with unequal magnitudes: Opposite charges. Note that twice the number of flux lines enter (or leave) the charge that is twice as large DEMOS: Plumes (E-9) and wheat + parallel plates # lines proportional to the magnitude of charge

Quiz 2 (lecture 2) Two charges q1 and q2 are fixed at points (-a,0) and (a,0) as shown. Together they produce an electric field at point (0,d) which is directed along the negative y-axis. y (0,d) Both Charges are negative q1 q2 (-a,0) (a,0) x Which of the following statements is true: A) Both charges are negative B) Both charges are positive C) The charges are opposite D) There is not enough information to tell how the charges are related

Electric Field From Two Charges + _ _ +

Electric Field due to an Electric Dipole From symmetry must lie on dipole axis + - P Dipole center z r+ r- q+ q- d Eminus points downward because we are using a positive test charge at point P and the positive test charge is attracted to the negative end of the dipole. Use Binomial Expansion & d << z, the magnitude of the electric field at P:

Electric Field due to an Electric Dipole + - P Dipole center z r+ r- q+ q- The electric dipole moment (C-m): + Direction of p is from negative to positive d - Rewrite magnitude of electric field at P:

Electric Dipole Moment L is the separation of the two charges Dipole moment: The direction of the dipole moment is taken to be from the negative to the positive end of the dipole. p 610-30 C·m0.04 e·nm

Electric Dipole in a uniform electric field Because the field is uniform, the net force is zero. However, there is a net torque.

Electric dipole in a uniform electric field DEMO 5A-31 When q = 0o or 180o,  = 0. However,  = 180o is unstable. For clockwise rotation

A Point Charge in a Uniform Field An particle of charge q injected in a E field will experience a force given by F = q E. The resulting acceleration can be found from Newton's second law. In a region of uniform E field, the particle will experience a constant acceleration and the resulting parabolic trajectory. The control of electrons by so-called deflection plates is the principle behind the operation of the cathode-ray tube used in oscilloscopes and many televisions and computer monitors. DEMO

Point Charge in Uniform E field - - - - - - - - - - - - - - - - - - + + ++ + + + + + + ++ + q E v0 x

Point Charge in Uniform E field - - - - - - - - - - - - - - - - - - + + ++ + + + + + + ++ + P E v0 x If E=100N/C v0=5105m/s x=0.5 m Then

Quiz 3 (lecture 2) Three point charges are arranged at the corners of a square as shown. What is the direction of the electric field at lower right corner of the square at x? (Take Q > 0.) +Q - Q toward the lower right toward the upper left Upward to the right none of the above +Q

Quiz 4 (lecture 2) In which of the two cases shown below is the magnitude of the electric field at the point labeled A the largest? Case 1 Case 2 Equal +Q -Q A Case 1 Case 2 The upper left +Q only affects the x direction in both and the lower right +/-Q only affects the y direction….so in both, nothing cancels out, so they’ll have the same magnitude

Quiz 5 (lecture 2) +2Q +Q qo r 2r A positive test charge qo is released from rest at a distance r away from a charge of +Q and a distance 2r away from a charge of +2Q. How will the test charge move immediately after being released? +2Q +Q qo r 2r To the left To the right It doesn’t move at all. Moves upward Moves downward

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Quiz 6 (lecture 2) Examine the electric field lines produced by the charges in this figure. Which statements are true? q1 q2 (a) q1 and q2 have the same sign and | q1 | = | q2 | (b) q1 >0 and q2 <0 and | q1 | < | q2 | (c) q1 <0 and q2 >0 and | q1 | < | q2 | Field lines start from q2 and terminate on q1. This means q2 is + ; q1 is - Now, which one is bigger? More lines leave q2 than enter q1. So q2 has more charge!

Potential Energy of an Electric Dipole Potential energy can be associated with the orientation of an electric dipole in an electric field. U is least =0 U=-pE U is greatest =180 U=pE U =0 when =90

Electric Field With this concept, we can “map” the electric field anywhere in space produced by any arbitrary: Bunch of Charges Charge Distribution F Using principal of superposition “Net” E at origin + - + + These charges or this charge distribution “source” the electric field throughout space

Electric Field due to an Electric Dipole + - P Dipole center z r+ r- q+ q- In general: + - *Where r is the distance between the dipole center and the point in question. *regardless of wherever the point lies on the dipole axis. d

Quiz 7 (lecture 2) The magnitude and direction of the electric field is represented by the blue arrows. Which of the diagrams best represents the electric field from a negative charge?

Quiz 8 (lecture 2) y P y P ? An electric dipole is oriented along the y-axis as shown. You are not told which charge is positive, but the electric field at point P on the y-axis is shown by the red arrow. If the bottom charge is removed, which of the following best describes what happens to the electric field at P? Remove bottom charge The magnitude decreases and the direction stays the same The magnitude increases and the direction stays the same The magnitude stays the same but the direction changes The magnitude decreases and the direction changes There is not enough information to tell

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