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Chapter 22

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  • Chapter 22:Coulombs Law and Electric Fields Part IObjectivesTo study quantitatively the nature of forces between electric chargesTo develop the concept of the electric field as a transmitter of electric forceTo learn how work is done and potential energy is stored in an electric fieldTo learn how to use electric fields to control and direct the motion of electric charges

  • Coulombs LawExperiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them.Rye High School: Physics*

    Physics is Life

  • Coulombs LawCoulombs law:Where Q1 and Q2 are the amount of charge and k is a proportionality constantCharges produced by rubbing ordinary objects (such as a comb or a plastic ruler) are typically around a microcoulomb or less:Rye High School: Physics*

  • *Coulombs LawThe magnitude of the charge of an electron, on the other hand, has been determined to be about 1.602 x 10-19 C, and its sign is negative. This is the smallest known charge, and because of its fundamental nature, it is given the symbol e and is often referred to as the elementary charge:Example:How many electrons make up a charge of -30.0 micro coulombs (C)?N = Q/e = (-30 x 10-6 C)/ (-1.60 x 10-19 C/electrons) = 1.88 x 1014 electronsWhat is the mass of 1.88 x 1014 electrons?Mass = (9.11 x 10-31 kg)(1.88 x 1014 electrons) = 1.71 x 10-16 kgRye High School: Physics

  • The charges carried by the proton and electron are equal in size. However, the mass of the proton is 2000 times the mass of the electron.Coulombs LawRye High School: Physics*

  • Double one of the chargesforce doublesChange sign of one of the chargesforce changes directionChange sign of both chargesforce stays the sameDouble the distance between chargesforce four times weakerDouble both chargesforce four times strongerCoulombs LawRye High School: Physics*

  • Coulombs Law vs. Law of Universal GravitationF = kQ1Q2/r2 vs. F=GM1M2/r2Both are inverse square laws F1/r2Both have a proportionality to a product of each body-mass for gravity, electric charge for electricity.A major difference is that gravity is always an attractive force, whereas the electric force can be wither attractive or repulsive.Electrical Force is stronger than gravitational force

    Rye High School: PhysicsComparison of electrical force vs. Gravitational force #1

    Comparison of electrical force vs. Gravitational Force #2*

  • *Sample problemFind the force between two positive 1.0 C charges when they are 1000m apart?Solutionq1=q2 = 1.0C r = 1000mF = kq1q2/r2 where k = 9.0 x 109 Nm2/C2After substitution, F = 9.0 x 103 NSolving Problems involving Coulombs LawRye High School: Physics

  • Solving Problems involving Coulombs LawSample problemWhat is the magnitude of the electric force of attraction between an iron nucleus (q = +26e) and its innermost electron if the distance between them is 1.5 x 10-12 m?SolutionF = kq1q2/r2 where k = 9.0 x 109 Nm2/C2F = (9.0 x 109 Nm2/C2)(26)(1.6 x 10-19 C)(-1.6 x 10-19 C)/ (1.5 x 10-12 m)2 = -2.7 x 10-3 N Rye High School: Physics*

  • Solving Problems involving Coulombs LawSample problemWhat is the repulsive electrical force between two protons in a nucleus that are 5.0 x 10-15 m apart from each other?SolutionF = kq1q2/r2 where k = 9.0 x 109 Nm2/C2F = (9.0 x 109 Nm2/C2)(1.6 x 10-19 C)(1.6 x 10-19 C)/ (5.0 x 10-15 m)2 = 9.2 N Rye High School: Physics*

  • Solving Problems involving Coulombs LawSample problemTwo charged balls are 20.0 cm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?SolutionLet F1 = kq1q2/r12 and F2 = kq1q2/r22 where F2 = 3 F1F2/F1 = r12 /r223= [(20.0cm)/r2]2, which gives r2 = 11.5 cmRye High School: Physics*ADVANCEDADVANCED

  • *In physics, the space surrounding an electric charge has a property called an electric field. This electric field exerts a force on other electrically charged objects. The concept of an electric field was introduced by Michael Faraday.

    The electric field is a vector field with SI units of newtons per coulomb (N C1) or, equivalently, volts per meter. The strength of the field at a given point is defined as the force that would be exerted on a positive test charge of +1 coulomb placed at that point; the direction of the field is given by the direction of that force. The Electric FieldMichael Faraday (1791-1867)Rye High School: Physics

  • *One can think of electric force as establishing a field telling particles which way to move and how fast+Electric field lines tell a positivecharge which way to move.

    For example, a positive charge itselfhas field lines pointing away from it,because this is how a positively-chargedtest-particle would respond if placedin the vicinity (repulsive force). +Run Away!The Electric FieldRye High School: Physics

  • The Electric Field++++The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source chargeRye High School: Physics*

  • *Measuring an electric field is a quite simple process involving a test charge. To measure the strength of an electric field, first a test charge must be placed in its vicinity, then calculate the force the test charge feels. The resulting number is the strength of the electric field. This process is simplified into the following equation

    In this equation,F is the magnitude of the force, as found by using Coulomb's Law, q is the magnitude of the test charge. The resulting electric strength is measured in Newtons per a Coulomb. The Electric FieldRye High School: Physics

  • *Electric Field vs. Gravitational FieldRight now you are experiencing a uniform gravitational field: it has a magnitude of 9.8 m/s2 and points straight down. If you threw a mass through the air, you know it would follow a parabolic path because of gravity. You could determine when and where the object would land by doing a projectile motion analysis, separating everything into x and y components. The horizontal acceleration is zero, and the vertical acceleration is g. We know this because a free-body diagram shows only mg, acting vertically, and applying Newton's second law tells us that mg = ma, so a = g. You can do the same thing with charges in a uniform electric field. If you throw a charge into a uniform electric field (same magnitude and direction everywhere), it would also follow a parabolic path. We're going to neglect gravity; the parabola comes from the constant force experienced by the charge in the electric field. Again, you could determine when and where the charge would land by doing a projectile motion analysis. The acceleration is again zero in one direction and constant in the other. The value of the acceleration can be found by drawing a free-body diagram (one force, F = qE) and applying Newton's second law. This says: qE = ma, so the acceleration is a = qE / m. The one big difference between gravity and electricity is that m, the mass, is always positive, while q, the charge, can be positive, zero, or negative. Rye High School: Physics

  • *The Electric FieldSample Problem A positive charge of 1.0 x 10-5C experiences a force of 0.30N when located at a certain point. What is the electric field intensity at that point?SolutionE=F/q = 0.30N / 1.0 x 10-5 C = 3.0 x 104 N/CRye High School: Physics

  • Sample Problem A test charge experiences a force of 0.20 N on it when it is placed in an electric field intensity of 4.5 x 105 N/C. What is the magnitude of the charge?Solutionq=F/E = 0.20N / 4.5 x 105 N/C = 4.4 x 10-7 CRye High School: PhysicsThe Electric Field*

  • Sample ProblemA positive charge of 10-5 C experiences a force of 0.2N when located at a certain point in an electric field. What is the electric field strength at that point?SolutionF= 0.2N q=10-5CE= F/q = 0.2N/10-5C = 2 x 104 N/CRye High School: PhysicsThe Electric Field*

  • Rye High School: PhysicsMeasuring Potential Difference (Voltage)

    Electric Strength can also be defined in units of work and energy. This alternative way of measuring the strength of an electric field entails finding the potential difference, which exits between any two points. To visualizing this difference picture an electric field, then trying to force a test charge in between two points, if the test charge is repelled, we have to do work to push it into place. This process is also simplified into the formula

    In this formula, W is work, q is the magnitude of charge, and V is voltage or potential difference. *

  • Rye High School: PhysicsMeasuring Potential Difference (Voltage)Sample ProblemIt takes 5.0 x 10-3 J of work to move a positive charge of 2.5 x 10-4C from point X to point Y on an electric field. What is the difference of Potential between X and Y?

    SolutionW= 5.0 x 10-3 J q = +2.5 x 10-4CV = W/q = 5.0 x 10-3J/2.5 x 10-4C = 20 J/C = 20 volts*

  • Rye High School: PhysicsElectronvoltThe joule is too large a unit for measuring the work done on moving elementary charges, such as electrons, protons, or the small charges in ions. For this purpose, the electron volt is a more convenient unit of energy or work. An electron volt is the work done in moving an electron or other body having a unit of elementary charge through a difference of potential of one volt. Thus,1eV = 1.60 x 10-19 J*

  • Rye High School: PhysicsElectronvoltSample ProblemThe difference of potential between point X and point Y of an electric field is 100 volts. (a) How much work done in electronvolts is done by the electric field in moving a free electron from point X and Y. (b) What happens to this work? (c) What is this