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Announcements Jan 11 • Yes the University is open all day (?) • Webassign Online forum schedule: – Fri, Sat, Sun, Mon. evenings 7-10pm before a homework is due on a Tuesday morning – See web site • Help labs have started – See course web page for details http://www.physics.utah.edu/~woolf/2020_Jui/help_lab.pdf 1 18.5 Coulomb’s Law Example A Model of the Hydrogen Atom In the Bohr model of the hydrogen atom, the electron is in orbit about the nuclear proton at a radius of 5.29x10-11m. Determine the speed of the electron, assuming the orbit to be circular. 2 18.5 Coulomb’s Law Example A Model of the Hydrogen Atom In the Bohr model of the hydrogen atom, the electron is in orbit about the nuclear proton at a radius of 5.29x10-11m. Determine the speed of the electron, assuming the orbit to be circular. The force on the electron is exerted by the proton, as given by Coulomb’s Law F =k q1 q2 r2 = (8.99 ×10 9 )( N ⋅ m C 1.60 ×10 2 2 (5.29 ×10 −11 m v= Fr = m C ) 2 2 = 8.22 ×10 −8 N mv 2 F = mac = r But the SAME force is responsible for the centripetal motion Solving for v gives ) −19 (8.22 × 10 N )(5.29 × 10 −8 9.11 × 10-31 kg −11 m) = 2.18 × 106 m s 3 18.5 Coulomb’s Law Example Three Charges on a Line Determine the magnitude and direction of the net force on q1. 4 18.5 Coulomb’s Law Example Three Charges on a Line Determine the magnitude and direction of the net force on q1. qi q j Fij = Fij = k 2 rij where rij represents the distance between qi and F12 = k q1 q2 F13 = k q1 q3 r12 r13 2 2 qj (8.99 × 10 = (8.99 × 10 = 9 9 N ⋅ m 2 C2 )(3.0 × 10−6 C )(4.0 × 10−6 C (0.20m )2 N ⋅ m 2 C2 )(3.0 × 10−6 C )(7.0 × 10−6 C (0.15m )2 ) ) = 2.7 N = 8.4 N F1 = F12 + F13 = −2.7 Nˆi + 8.4 Nˆi = + 5.7Nˆi Here ˆi denotes the unit vector in the + x direction 5 Example Three Charges not in line Assuming the horizontal to be the x-direction, and vertical the y-direction, determine the magnitude and direction of the net force on q1 6 Example Three Charges not in line Assuming the horizontal to be the x-direction, and vertical the y-direction, determine the magnitude and direction of the net force on q1 7 18.5 Coulomb’s Law Example Three Charges not in line Assuming the horizontal to be the x-direction, and vertical the y-direction, determine the magnitude and direction of the net force on q1 Solution: First we apply the Inverse Square (Coulomb’s) Law to find the magnitudes of the two forces on q1 : qi q j Fij = Fij = k 2 rij where rij represents the distance between qi and F12 = k q1 q2 F13 = k q1 q3 r12 r13 2 2 qj (8.99 × 10 = (8.99 × 10 = 9 9 N ⋅ m 2 C2 )(4.0 × 10−6 C )(6.0 × 10−6 C (0.15m )2 ) N ⋅ m 2 C2 )(4.0 × 10−6 C )(5.0 × 10−6 C (0.10m )2 ) = 9.59 N = 17.98 N 8 18.5 Coulomb’s Law The resulting force F on q1 is the vector sum of the forces exrted by charges q2 and q3. We make this sum component by component: Both forces are attractive here, with dirrections indicated in the figure to the left Fx = + F12 cos θ 2 + F13 cos θ 3 = (9.59 N ) cos(73o ) + (17.95 N ) cos(0o ) = 2.80 N + 17.95 N = 20.75 N Fy = + F12 sin θ 2 + F13 sin θ 3 = (9.59 N ) sin(73o ) + (17.95 N ) sin(0o ) = 9.17 N + 0 = 9.17 N F = Fx2 + Fy2 = ( 20.75 N ) 2 + (9.17 N ) 2 = 23 N Angle of F counter - clock - wise from from the + x - axis : θ = tan -1 Fy Fx = tan −1 (0.442) = 24o 9 18.6 The Electric Field DEFINITION OF ELECRIC FIELD The electric field that exists at a point is the electrostatic force experienced by a (small) test charge q0 placed at that point divided by the charge q0 itself: F E= qo SI Units of Electric Field: newton per coulomb (N/C) NOTE: It is the surrounding charges that create the electric field at a given point. The effect of the test charge itself is NEVER included in this definition 10