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Coulomb’s Law Physics 12 Electroscope ► Draw a diagram showing the forces that exist on the leaves of the electroscope when it is placed close to the Van de Graff generator Coulomb’ Coulomb’s Pendulum ► Consider the following pendulum ► The mass initially shares charge with an identical sphere ► The two spheres then have identical charge and (in this case) the second sphere is moved toward the pendulum bob from the left, causing the pendulum to deflect 1 Coulomb’ Coulomb’s Pendulum ► Draw a free body diagram for the pendulum bob; you will know the mass of the bob and the separation of the two spheres and the angle the pendulum support makes with the vertical ► Develop an equation that will allow you to solve for the electrostatic force experienced by the pendulum (and the second sphere) Pith Ball Fe − Tx = 0 Fe = Tx Fe = T cos θ mg sin θ cos θ Fe = mg sin θ mg Fe = tan θ T= Fg − Ty = 0 Fg = Tg Fg = T sin θ Coulomb’ Coulomb’s Pendulum Separation (m) Angle (with vertical) 0.150 11.52319 0.130 15.18589 0.110 20.76193 0.090 29.52358 0.070 43.11136 2 Coulomb’ Coulomb’s Pendulum ► What type of relationship does this display as the distance between the point charges is increased or decreased? Coulomb’ Coulomb’s Law ► Coulomb’ Coulomb’s Law is given as: Fe = kq1q2 r2 Electric Force: Coulomb’ Coulomb’s Law ► Electric Force is the force felt by separated (positive or negative) charges. Opposite charges attract Like charges repel Fe = kq1q2 r2 + - + + k = 9.0 ×109 [ ] Nm 2 C2 3 Coulomb’ Coulomb’s Law ► Positive force will result when: Two positively charged particles Two negatively charged particles ► Negative Fe = kq1q2 r2 force will result when: One positively and negatively charged particle ► Positive force indicates repulsion force indicates attraction ► Negative Practice ► What attractive force does an electron in a hydrogen atom experience? ► What attractive force does an electron in helium experience? Applications of Coulomb’s Law Physics 12 4 Coulomb Sample Problem ► Three charges are arranged in a line; if the three charges are 15µ 15µC, -12µ 12µC and 18µ 18µC respectively. The distance between the first two charges is 0.20m and the second and third charges is 0.30m. What is the force experienced by the first charge? Coulomb Sample Problem ► What force is experienced by the remaining two charges? Force (Vector) Addition ► To add forces, resolve each force into its components and treat the forces in the xxdirection and yy-direction independently ► Once you sum the x and y components, use Pythagorean Theorem and Trigonometry to resolve into a resultant force 5 Example ►A point P has forces of 12.0N at 24.3° 24.3°, 17.6N at 112° 112°, 6.78N at 241° 241° and 10.2N at 74.4° 74.4°. Determine the resultant vector ► 25.5N, 81.4° 81.4° Coulomb’ Coulomb’s Law and Vector Addition ► When we consider an electrostatic system, we need to use Coulomb’ Coulomb’s Law to determine the magnitude and direction of each force ► Once the magnitude and direction of each force has been determined, then the vector sum can be completed Coulomb’ Coulomb’s Law in 2D ► Three charges are arranged as follows; a 2.0µ 2.0µC is placed 4.0m due north of a 3.0µ 3.0µC charge and 3.0m due west of a 5.0µ 5.0µC charge. What is the force experienced by the -2.0µ 2.0µC charge? 6 Coulomb’ Coulomb’s Law in 2D 1 -2.0µC 4.0m 2 3.0 µ C 3.0m 5.0 µ C 3 kq q Fe12 = 12 2 r Fe12 = −3.4x10−3 N kq3q2 r2 Fe23 = 5.4x10−3 N Fe23 = Coulomb’ Coulomb’s Law in 2D θ = 53o + 180o 2 3.0 µ C θ = 233o Fe12 x = 0 N Fe12 y = 3.4 x10 −3 N Fe 23 x = 5.4 x10 −3 N cos 233 Fe 23 x = −3.2 x10 −3 N Fe 23 y = 5.4 x10 −3 N sin 233 Fe 23 y = −4.3 x10 −3 N Coulomb’ Coulomb’s Law in 2D ► Use the x and y component data to determine the resultant force vector Fex = 0 N − 3.2 x10 −3 N Fe = 3.3x10 −3 N Fex = −3.2 x10 −3 N θ = 16o Fey = 3.4 x10 −3 N − 4.3x10 −3 N r Fe = 3.3x10 −3 N ,196o Fey = −9.0 x10 − 4 N 7 Electric Fields Physics 12 Field Theory ► When forces exist without contact, it can be useful to use field theory to describe the force experienced by a particle at any point in space ► We live in a gravitational field where the separation between massive objects results in attractive forces ► In a similar way, we can think of an electric field Electric Field Mapping ► To map an electric field, a small positive test charge is placed in the field and the magnitude and direction of the force is recorded ► The test charge is then moved throughout the electric field and a map of the field is created ► If the force experienced by the test charge can be measured, then we can map the field 8 Test Charge ► The test charge that is used must be small compared to the charge creating the field ► If not, the test charge’ charge’s field will change the field that is being investigated ► The electric field should be the same regardless of the test charge used Field Lines – Two Positive Charges 9 Field Lines – Two Opposing Charges Problem ► What are the relative magnitudes of the charges in the diagram? ► What is the polarity of each of the charges? Multiple Charges ► It is also possible to consider what happens with multiple charges: 10 Electric Field Intensity ► The electric field can be determined using the force experienced by a particle and the charge on the particle r Fonq ' E = q' Electric Field ► The electric field also has a direction; since the field is the superposition of all the electric field vectors at a given point in space where: v r Fonq '1 E1 = q' r r E = ΣEi Electric Field ►A charge of 2.0mC is placed at the origin and a charge of -5.0mC is placed at the point (3,0); what electric field exists at: (1,0) (4,0) (-1,0) ► Where is the electric field equal to zero? 11 Electric Potential Physics 12 Electric Potential Energy ► Gravitational potential energy is due to mass, gravitational field intensity and separation ► Electric potential energy is due to charge, electric field intensity and separation E p = Eqr F qr q kqQ Ep = 2 r r kqQ Ep = r Ep = Potential Difference ► Similar to gravitational potential difference, electric potential difference is measured with respect to a reference point (usually the ground) which we call zero ► This concept is not as useful for gravitational difference as objects have different masses, but since each charge carrier has the same charge, this concept has value for electric potential difference 12 Voltage or Potential Difference ► Electric potential difference is known as voltage ► One volt is defined as one joule per coulomb ► Electric Potential is NOT electric potential energy V = Ep q kQ V = r These plots show the potential due to (a) positive and (b) negative charge. What minimum work is required by an external force to bring a charge q = 3.00 µC from a great distance away (r = infinity) to a point 0.500 m from a charge Q = 20.0 µC ? 13 Analogy between gravitational and electrical potential energy: Both rocks have the same gravitational potential, but the bigger rock has more Ep. Both charges have the same electric potential, but the 2Q charge has more Ep. What is the change in potential energy of the electron in going from a to b? What is the speed of the electron as a result of this acceleration? Repeat both calculations for a proton. Uniform Electric Field ► As previously seen, it is possible to produce a uniform electric field ► The intensity of the field is a function of the voltage and separation of the plates V = Ed 14 Equipotential Lines •An equipotential (represented by the green dashed lines) is a line or surface over which the potential is constant. •Electric field lines are perpendicular to equipotentials. •The surface of a conductor is an equipotential. Equipotential Lines The Electron Volt, a Unit of Energy A Joule is too large when dealing with electrons or atoms, so electron volts are used. One electron volt (eV) is the energy gained by an electron moving through a potential difference of one volt. 15 Elementary Charge ► Robert Millikan investigated the charge on an electron in his famous oiloil-drop experiment ► He was awarded the Nobel Prize in 1923 for his 1917 research that led to the elementary charge of 1.60x10-19C ► Today, the accepted value of the elementary charge is 1.60217733x10-19C Elementary Charge ► Since we know the value of the elementary charge, we can determine the number of charge carriers or the total charge with the following equation q = Ne 16