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Transcript
Lecture 2 1 Wednesday, January 15, 2014 iClicker question You place same two charges a distance r apart. Then you double each charge and double the distance between the charges. How does the force between the two charges change? a) The new force is twice a large. b) The new force is half as large. c) The new force is four times as large. d) The new force is four times smaller. e) The new force is the same 2 Wednesday, January 15, 2014 Electric Force - Coulomb’s Law Consider two electric charges: q1 and q2 The electric force F between these two charges separated by a distance r is given by Coulomb’s Law The constant k is called Coulomb’s constant and is given by Coulomb’s law in vector form q1 q2 r12 F=k 2 r12 r12 32 Wednesday, January 15, 2014 2 2 2 q1 q2 q (2q) q F =k 2 =k 2 →k = k 2 2 r r (2r) r Force remains the same 4 Wednesday, January 15, 2014 Demo: electrostatic charging 5 Wednesday, January 15, 2014 +++ --- 6 Wednesday, January 15, 2014 F +++ --- 6 Wednesday, January 15, 2014 F +++ +++ --- --- 6 Wednesday, January 15, 2014 F +++ +++ --F - --- 6 Wednesday, January 15, 2014 F +++ +++ +++ + --F - ----- 6 Wednesday, January 15, 2014 F +++ --F +++ - --- F +++ + --- 6 Wednesday, January 15, 2014 Charging by induction --++ ++ +q + + extra electrons +q + + This will not work if no free electrons + + ++ + ++-+- ++ -+ - + - +q 7 Wednesday, January 15, 2014 Example - Forces between Electrons What is relative strength of the electric force compared with the force of gravity for two electrons? Gravity is irrelevant for atomic and subatomic processes – the electric force is much much stronger. But sometimes gravity is most important; e.g., the motion of the planets. 52 Wednesday, January 15, 2014 Example - Forces between Electrons What is relative strength of the electric force compared with the force of gravity for two electrons? Gravity is irrelevant for atomic and subatomic processes – the electric force is much much stronger. But sometimes gravity is most important; e.g., the motion of the planets. Why? If Coulomb forces are 42 orders of magnitude stronger than gravity, why we do not “clearly” feel it? 52 Wednesday, January 15, 2014 Superposition Principle The net force acting on any charge is the linear vector sum of the forces due to the remaining charges in the distribution. 37 Wednesday, January 15, 2014 Vector addition of force Electric force in vector form F23 q1 Ftot=F13+F23 r13 r23 q3 F13 q2 10 Wednesday, January 15, 2014 iClicker question Three charges are arranged on a straight line as shown in the figure. What is the direction of the electrostatic force on the middle charge? Wednesday, January 15, 2014 x Positive force directed along x, negative along “- x” F32 1 a 2 a F32 2 q = k 2 = F12 a 3 F12 12 Wednesday, January 15, 2014 x Positive force directed along x, negative along “- x” F32 1 a 2 a F32 2 q = k 2 = F12 a 3 F12 12 Wednesday, January 15, 2014 iClicker question Three charges are arranged on a straight line, separated by “a”. What is the force on the right charge? (Positive: along x) x a a 2 q A : 0; B : −k 2 ; 2a 2 2 q q D : −k 2 ; E: k 2 a a Wednesday, January 15, 2014 2 q C: k 2 2a iClicker question 1 a x a 2 2 3 F23 F13 2 F13 2 (−2)q q =k = −k 2 2 4a 2a 2 Ftot = F13 + F23 Wednesday, January 15, 2014 q =k 2 a 2 2 q q q = −k 2 + k 2 = k 2 2a a 2a Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? 43 Wednesday, January 15, 2014 Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? 43 Wednesday, January 15, 2014 General Problem Solving Strategies 1. Write what is given. 2. Sketch 3. Write equations 4. Count # of equations and unknown (should be equal!) 5. Solve the system (DO NOT PLUG IN NUMBERS!!!) 6. Check dimensions (if dimensions are wrong, the answer is wrong), Check simple cases 7. Plug in the numbers... University Physics, Chapter 1 Wednesday, January 15, 2014 29 General Problem Solving Strategies 1. Write what is given. 2. Sketch 3. Write equations 4. Count # of equations and unknown (should be equal!) 5. Solve the system (DO NOT PLUG IN NUMBERS!!!) 6. Check dimensions (if dimensions are wrong, the answer is wrong), Check simple cases 7. Plug in the numbers... University Physics, Chapter 1 Wednesday, January 15, 2014 29 Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? 45 Wednesday, January 15, 2014 Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? 45 Wednesday, January 15, 2014 Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? Write what is given: 45 Wednesday, January 15, 2014 Example - Charged Balls Consider two identical charged balls hanging from the ceiling by strings of equal length ℓ = 1.5 m (in equilibrium). Each ball has a charge of q=25 µC. The balls hang at an angle θ = 25° with respect to the vertical. What is the mass of the balls? Write what is given: ℓ = 1.5 m, q1=q2=25 µC, θ = 25° 45 Wednesday, January 15, 2014 Sketch Three forces act on each ball: Coulomb force Fc, gravity Fg, and the tension of the string T The Coulomb force is horizontal and must be repulsive to keep the balls apart The gravitational force points down The tension T is in the direction of the string The balls are in equilibrium, which means the three forces are balanced Add the three forces to the drawing Define distance d between the balls 18 Wednesday, January 15, 2014 Example - Charged Balls Make a separate free body diagram for one of the charged balls Define x-y coordinate system 46 Wednesday, January 15, 2014 Example - Charged Balls Make a separate free body diagram for one of the charged balls Define x-y coordinate system 46 Wednesday, January 15, 2014 Example - Charged Balls Write equations The condition of static equilibrium tells us that the sum of the x-components of the three forces acting on the ball must equal zero and the sum of y-components of these forces must equal zero The sum of the x-components of the forces is • T is magnitude of the string tension • θ is the angle of the string relative to the vertical • FC is the magnitude of the Coulomb force The sum of the y-components of the forces is The force of gravity, Fg, is just the weight of the charged ball • m is the mass of the charged ball 47 Wednesday, January 15, 2014 Example - Charged Balls Write equations The condition of static equilibrium tells us that the sum of the x-components of the three forces acting on the ball must equal zero and the sum of y-components of these forces must equal zero The sum of the x-components of the forces is • T is magnitude of the string tension • θ is the angle of the string relative to the vertical • FC is the magnitude of the Coulomb force The sum of the y-components of the forces is The force of gravity, Fg, is just the weight of the charged ball • m is the mass of the charged ball 47 Wednesday, January 15, 2014 Example - Charged Balls The electric force between the two balls is • d is the distance between the two balls We can express d in terms of ℓ and θ 48 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 2 q Fc = k 2 d d/2 sin θ = l 5=5 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg Fg = mg 5=5 2 q Fc = k 2 d d/2 sin θ = l } eliminate d 22 Wednesday, January 15, 2014 Count # of equations and unknown (should be equal!) T sin θ = Fc T cos θ = Fg } Fg = mg 5=5 2 q Fc = k 2 d d/2 sin θ = l eliminate T } eliminate d 22 Wednesday, January 15, 2014 Example - Charged Balls Solve the system We divide the two free-body force equations containing the string tension Thus we eliminate the string tension and get Putting in our expressions for the force of gravity and the electric force, we get 49 Wednesday, January 15, 2014 Example - Charged Balls Solve the system We divide the two free-body force equations containing the string tension Thus we eliminate the string tension and get Putting in our expressions for the force of gravity and the electric force, we get 49 Wednesday, January 15, 2014 Check dimensions Dimensions 2 2 Nm C kg = 2 2 2 C (m/s )m 2 N = kgm/s 2 2 2 (kgm/s )m C kg = 2 2 2 C (m/s )m 24 Wednesday, January 15, 2014 Check dimensions Dimensions 2 2 Nm C kg = 2 2 2 C (m/s )m 2 N = kgm/s 2 2 2 (kgm/s )m C kg = 2 2 2 C (m/s )m 24 Wednesday, January 15, 2014 Check dimensions Dimensions 2 2 Nm C kg = 2 2 2 C (m/s )m 2 N = kgm/s 2 2 2 (kgm/s )m C kg = 2 2 2 C (m/s )m 24 Wednesday, January 15, 2014 Check dimensions Dimensions 2 2 Nm C kg = 2 2 2 C (m/s )m 2 N = kgm/s 2 2 2 (kgm/s )m C kg = 2 2 2 C (m/s )m 24 Wednesday, January 15, 2014 Check dimensions Dimensions 2 2 Nm C kg = 2 2 2 C (m/s )m 2 N = kgm/s 2 2 2 (kgm/s )m C kg = 2 2 2 C (m/s )m 24 Wednesday, January 15, 2014 Simple cases? Large mass -> small angle: OK Large charge-> large mass (for given angle): OK Shorter line: large angle: OK 25 Wednesday, January 15, 2014 Simple cases? Large mass -> small angle: OK Large charge-> large mass (for given angle): OK Shorter line: large angle: OK 25 Wednesday, January 15, 2014 Calculate Now, put in numbers... Putting in our numerical values we obtain Round 50 Wednesday, January 15, 2014 Calculate Now, put in numbers... Putting in our numerical values we obtain Round 50 Wednesday, January 15, 2014 Calculate Now, put in numbers... Putting in our numerical values we obtain Round 50 Wednesday, January 15, 2014 Calculate Now, put in numbers... Putting in our numerical values we obtain Round We report our result to three significant figures 50 Wednesday, January 15, 2014 Electric field 27 Wednesday, January 15, 2014 The Electric Field A charge creates an electric field around itself and the other charge feels that field. + Electric field at a given point in space: place a positive test charge q at the point and measure the electrostatic force that acts on the test charge; then 3 Wednesday, January 15, 2014 The Electric Field A charge creates an electric field around itself and the other charge feels that field. + + Electric field at a given point in space: place a positive test charge q at the point and measure the electrostatic force that acts on the test charge; then Test charge q 3 Wednesday, January 15, 2014 The Electric Field A charge creates an electric field around itself and the other charge feels that field. + + Electric field at a given point in space: place a positive test charge q at the point and measure the electrostatic force that acts on the test charge; then Test charge q 3 Wednesday, January 15, 2014 Precise Definition of Electric Field We define the electric field in terms of the force it exerts on a positive point charge: Unit of the electric field: N/C (Newtons per Coulomb) Note that the electric force is parallel to the electric field and is proportional to the charge • Force alined with field for q>0, counter alined for q< 0 E 5 Wednesday, January 15, 2014 Precise Definition of Electric Field We define the electric field in terms of the force it exerts on a positive point charge: Unit of the electric field: N/C (Newtons per Coulomb) Note that the electric force is parallel to the electric field and is proportional to the charge • Force alined with field for q>0, counter alined for q< 0 E q>0, F 5 Wednesday, January 15, 2014 Precise Definition of Electric Field We define the electric field in terms of the force it exerts on a positive point charge: Unit of the electric field: N/C (Newtons per Coulomb) Note that the electric force is parallel to the electric field and is proportional to the charge • Force alined with field for q>0, counter alined for q< 0 q<0, F E q>0, F 5 Wednesday, January 15, 2014 EM fields are real physical substance Electromagnetic fields are real, these are not just mathematical concepts. EM fields is another material substance: one can assign energy-density to electromagnetic field. Empty space with EM field has more energy than just empty space. There is no “substance” that “transports” EM forces (like mechanical forces are transported by the medium, sound is transported by medium), EM wave propagate “on their own”, no medium to carry. (Actually, gravitational force is not “real”). 30 Wednesday, January 15, 2014 Example: Field of a Point Charge Question: What is the field created by a point charge q? Answer: Consider a “test charge” q0 at point x. Force on q0: E Electric field at x 6 Wednesday, January 15, 2014 Superposition of Electric Fields Suppose we have many charges The electric field at any point in space will have contributions from all the charges. The electric field at any point in space is the superposition of the electric field from n charges is (vectors!) Note that the superposition applies to each component of the field (x, y, z) 7 Wednesday, January 15, 2014 Vector addition of force Electric force in vector form F23 q1 Ftot=F13+F23 r13 r23 q3 F13 q2 33 Wednesday, January 15, 2014 Vector addition of force Electric force in vector form Еtot=Е1+Е23 Е2 q1 r13 r23 q3 Е1 q2 34 Wednesday, January 15, 2014 Electric Field Lines We can represent the electric field graphically by electric field lines — i.e., curves that represent the vector force exerted on a positive test charge. Electric field lines will originate on positive charges and terminate on negative charges. Electric field lines do not cross. • Why? Field lines are not real, just a useful mathematical concept 8 Wednesday, January 15, 2014 Electric Field Lines We can represent the electric field graphically by electric field lines — i.e., curves that represent the vector force exerted on a positive test charge. Electric field lines will originate on positive charges and terminate on negative charges. Electric field lines do not cross. • Why? Field lines are not real, just a useful mathematical concept 8 Wednesday, January 15, 2014 Electric Field Lines The electric force at a given point in space is tangent to the electric field line through that point. F2 F1 8 Wednesday, January 15, 2014 Electric Field Lines The electric force at a given point in space is tangent to the electric field line through that point. F2 F1 8 Wednesday, January 15, 2014 Properties of Field Lines The strength of the electric field is represented by the density of electric field lines Weak Strong The direction of the electric field is tangent to the electric field lines 9 Wednesday, January 15, 2014 Field Lines from a Point Charge The electric field lines from a point charge extend out radially For a positive point charge, the field lines point outward F • Terminate at infinity For a negative charge, the field lines point inward • Originate at infinity q E = k 2 r̂ r 10 Wednesday, January 15, 2014 Field Lines from a Point Charge The electric field lines from a point charge extend out radially For a positive point charge, the field lines point outward F • Terminate at infinity For a negative charge, the field lines point inward • Originate at infinity q E = k 2 r̂ r 10 Wednesday, January 15, 2014 Electric Field Lines for Two Point Charges We can use the superposition principle to calculate the electric field from two point charges. The field lines will originate from the positive charge and terminate on the negative charge. 3d 2d 11 Wednesday, January 15, 2014 Electric Field Lines (2) + - What is the direction of the E field at points halfway between two charges of different sign? Use superposition principle: 12 Wednesday, January 15, 2014 Electric Field Lines 13 Wednesday, January 15, 2014 iClicker A negative charge -q is placed in a nonuniform electric field as shown in the figure. What is the direction of the electric force on this negative charge? Wednesday, January 15, 2014 F=q E C since q< 0 E-field tangent to field line 43 Wednesday, January 15, 2014 End 44 Wednesday, January 15, 2014