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Electric Charges, Forces, and Fields

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Electric Charges, Forces, and Fields Electric Charges, Forces, and Fields Physics 231 Lecture 1-1 Fall 2008 Electric Charges Electric charge is a basic property of matter Two basic charges Positive and Negative Each having an absolute value of 1.6 x 10-19 Coulombs Experiments have shown that Like signed charges repel each other Unlike signed charges attract each other For an isolated system, the net charge of the system remains constant Charge Conservation Physics 231 Lecture 1-2 Fall 2008 Two basics type of materials Conductors Materials, such as metals, that allow the free movement of charges Insulators Materials, such as rubber and glass, that don’t allow the free movement of charges Physics 231 Lecture 1-3 Fall 2008 Coulomb’s Law Coulomb found that the electric force between two charged objects is Proportional to the product of the charges on the objects, and Inversely proportional to the separation of the objects squared q q F = k 1 2 r 2 k being a proportionality constant, having a value of 8.988 x 109 Nm2/c2 Physics 231 Lecture 1-4 Fall 2008 Electric Force As with all forces, the electric force is a Vector So we rewrite Coulomb’s Law as r q1q2 F12 = k rˆ12 r2 This gives the force on charged object 2 due to charged object 1 q1 q2 rˆ12 is a unit vector pointing from object 1 to object 2 The direction of the force is either parallel or antiparallel to this unit vector depending upon the relative signs of the charges Physics 231 Lecture 1-5 Fall 2008 Electric Force The force acting on each charged object has the same magnitude - but acting in opposite directions r r F12 = F21 (Newton’s Third Law) Physics 231 Lecture 1-6 Fall 2008 Example 1 A charged ball Q1 is fixed to a horizontal surface as shown. When another massive charged Q2 ball Q is brought near, it achieves an 2 Q2 equilibrium position at a distance d12 directly above Q1. d d 12 g 23 When Q1 is replaced by a different charged ball Q Q3, Q2 achieves an equilibrium position at a 1 Q3 distance d23 (< d12) directly above Q3. For 1a and 1b which is the correct answer 1a: A) The charge of Q3 has the same sign of the charge of Q1 B) The charge of Q3 has the opposite sign as the charge of Q1 C) Cannot determine the relative signs of the charges of Q3 & Q1 1b: A) The magnitude of charge Q3 < the magnitude of charge Q1 B) The magnitude of charge Q3 > the magnitude of charge Q1 C) Cannot determine relative magnitudes of charges of Q3 & Q1 Physics 231 Lecture 1-7 Fall 2008 Example 1 A charged ball Q is fixed to a horizontal surface 1 Q as shown. When another massive charged 2 ball Q2 is brought near, it achieves an Q2 equilibrium position at a distance d12 directly above Q . d d 1 12 g 23 When Q1 is replaced by a different charged ball Q1 Q3 Q3, Q2 achieves an equilibrium position at a distance d23 (< d12) directly above Q3. 1a: A) The charge of Q3 has the same sign of the charge of Q1 B) The charge of Q3 has the opposite sign as the charge of Q1 C) Cannot determine the relative signs of the charges of Q3 & Q1 • To be in equilibrium, the total force on Q2 must be zero. • The only other known force acting on Q2 is its weight. • Therefore, in both cases, the electrical force on Q2 must be directed upward to cancel its weight. • Therefore, the sign of Q3 must be the SAME as the sign of Q1 Physics 231 Lecture 1-8 Fall 2008 Example 1 A charged ball Q1 is fixed to a horizontal surface Q as shown. When another massive charged ball 2 Q2 Q2 is brought near, it achieves an equilibrium position at a distance d12 directly above Q1. d12 d23 When Q1 is replaced by a different charged ball g Q3, Q2 achieves an equilibrium position at a Q1 Q3 distance d23 (< d12) directly above Q3. 1b: A) The magnitude of charge Q3 < the magnitude of charge Q1 B) The magnitude of charge Q3 > the magnitude of charge Q1 C) Cannot determine relative magnitudes of charges of Q3 & Q1 • The electrical force on Q2 must be the same in both cases … it just cancels the weight of Q2 • Since d23 < d12 , the charge of Q3 must be SMALLER than the charge of Q1 so that the total electrical force can be the same!! Physics 231 Lecture 1-9 Fall 2008 More Than Two Charges Given charges q, q1, and q2 r If q1 were the only other charge, F q1 we would know the force on q q1q r due to q1 - Fq q r 1 q Fnet If q2 were the only other charge, r Fq q 2 we would know the force on q q2 due to q - r 2 Fq q 2 What is the net force if both charges are present? The net force is given by the Superposition Principle r r r Fnet = F1 +F 2 Physics 231 Lecture 1-10 Fall 2008 Superposition of Forces If there are more than two charged objects interacting with each other The net force on any one of the charged objects is The vector sum of the individual Coulomb forces on that charged object r qi Fj = q j ∑ k 2 rˆij i≠ j rij Physics 231 Lecture 1-11 Fall 2008 Example Two y (cm) q , q , and q are all point charges o 1 2 4 where qo = -1µC, q1 = 3µC, and 3 qo q2 = 4µC 2 1 What is the force acting on qo? q1 θ q2 r r r We have that F0 = F10 + F20 1 2 3 4 5 x (cm) What are F and F ? r r 0x 0y Need to calculate F10 and F20 r Decompose F 20 into its x and y q q r F = k 0 1 ˆ components 10 2 F10 = −F10 y r10 r F 20 = ()F 20 cos θ xˆ − (F 20 sin θ ) yˆ q0q2 r F = k F = −F rˆ x2 − x0 y0 − y2 20 2 20 20 20 cosθ = sinθ = r20 r20 r20 Physics 231 Lecture 1-12 Fall 2008 Example Two - Continued r r Now add the components of F 10 and F 20 to find F 0 x and F 0 y X-direction: F0x = F10x + F20x y (cm) 4 F10x = 0 3 q r o F F0x = F20 cosθ 20 2 r F 10 r 1 F0 q q Y-direction: 1 2 F0 y = F10 y + F20 y 1 2 3 4 5 x (cm) F0 y = −F10 − F20 sinθ Physics 231 Lecture 1-13 Fall 2008 Example Two - Continued Putting in the numbers . y (cm) 4 cosθ = 0.8 3 qo r F20 r = 5cm 2 r r10 = 3cm 20 F 10 r 1 F0 F10 = 30N F20 =14.4N q1 q2 1 2 3 4 5 x (cm) We then get for the components F0x =11.52N F0 y = −38.64N r 2 2 The magnitude of F 0 is F0 = F0x + F0 y = 40.32N At an angle given by −1 −1 o θ = tan (F0 y F0x )= tan (38.64/11.52) = −73.40 Physics 231 Lecture 1-14 Fall 2008 Note on constants k is in reality defined in terms of a more fundamental constant, known as the permittivity of free space. 1 k = 4πε0 2 −12 C with ε0 = 8.854x10 Nm2 Physics 231 Lecture 1-15 Fall 2008 Electric Field The Electric Force is like the Gravitational Force Action at a Distance The electric force can be thought of as being mediated by an electric field. Physics 231 Lecture 1-16 Fall 2008 What is a Field? A Field is something that can be defined anywhere in space A field represents some physical quantity (e.g., temperature, wind speed, force) It can be a scalar field (e.g., Temperature field) It can be a vector field (e.g., Electric field) It can be a “tensor” field (e.g., Space-time curvature) Physics 231 Lecture 1-17 Fall 2008 A Scalar Field 77 73 72 75 82 71 84 77 80 68 64 83 82 88 55 73 66 80 88 88 75 90 92 83 91 A scalar field is a map of a quantity that has only a magnitude, such as temperature Physics 231 Lecture 1-18 Fall 2008 A Vector Field 77 73 72 75 82 71 84 77 80 68 64 83 57 56 55 73 66 80 88 88 75 90 92 83 91 A vector field is a map of a quantity that is a vector, a quantity having both magnitude and direction, such as wind Physics 231 Lecture 1-19 Fall 2008 Electric Field We say that when a charged object is put at a point in space, The charged object sets up an Electric Field throughout the space surrounding the charged object It is this field that then exerts a force on another charged object Physics 231 Lecture 1-20 Fall 2008 Electric Field Like the electric force, the electric field is also a vector If there is an electric force acting on an object having a charge qo, then the electric field at that point is given by r r F E = (with the sign of q0 included) q0 Physics 231 Lecture 1-21 Fall 2008 Electric Field The force on a positively charged object is in the same direction as the electric field at that point, While the force on a negative test charge is in the opposite direction as the electric field at the point Physics 231 Lecture 1-22 Fall 2008 Electric Field A positive charge sets up an electric field pointing away from the charge A negative charge sets up an electric field pointing towards the charge Physics 231 Lecture 1-23 Fall 2008 Electric Field ⎛ ⎞ Earlier we saw that the ⎜ ⎟ r qi F = q ⎜ k rˆ ⎟ force on a charged object j j ⎜ ∑ 2 ij ⎟ ⎜ rij ⎟ is given by ⎝i≠ j ⎠ The term in parentheses remains the same if we change the charge on the object at the point in question The quantity in the parentheses can be thought of as the electric field at the point where the test object is placed The electric field of a point charge can then be shown to be given by r q E = k rˆ r 2 Physics 231 Lecture 1-24 Fall 2008 Electric Field As with the electric force, if there are several charged objects, the net electric field at a given point is given by the vector sum of the individual electric fields r r E = ∑ Ei i Physics 231 Lecture 1-25 Fall 2008 Electric Field If we have a continuous charge distribution the summation becomes an integral r dq E = k∫ rˆ r 2 Physics 231 Lecture 1-26 Fall 2008 Hints 1) Look for and exploit symmetries in the problem.
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