Millikan's Experiment Student Notes KEY PDF

# Physics 30 - Electrostatics – Lesson 8: Millikan's Experiment ## Particles in E and g - Recall that *F_e* is much smaller than *F_g* - Subatomic particles’ mass is very small, so when they are in both electric and gravitational fields, *F_net* is the net force changi...

# Physics 30 - Electrostatics – Lesson 8: Millikan's Experiment ## Particles in E and g - Recall that *F_e* is much smaller than *F_g* - Subatomic particles’ mass is very small, so when they are in both electric and gravitational fields, *F_net* is the net force changing their motion.                        - *10^-6 kg* 10^-9 kg - For “larger” masses (even mg or µg) *F_e* is a factor that needs to be considered. - Consider a negatively charged “larger” mass in a uniform electric field: - If it is accelerating: *F_net* = *F_e* + *F_g* - If it is suspended: *F_net* = 0 → *F_e* = *F_g* ## Example 1: An unknown charge is placed onto a 0.050 mg particle which is placed between two horizontal plates. The plates are 8.0 mm apart with a potential difference of 5 000 V across them. If the particle is suspended between the plates, what is the charge on the particle? How many excess electrons are on the particle? * d = 8.0 x 10^-3 m * m = 0.050 x 10^-6 kg * V = 5000 V *F_e* = *F_g* *qE* = *mg* *q* = *mgd* / *V* *q* = (0.050 x 10^-6 kg)(9.8 m/s²)(8.0 x 10^-3 m) / (5000 V) *q* = 7.848 x 10^-13 C Excess e: 7.848 x 10^-13 C / 1.6 x 10^-19 C = 4.9 x 10⁶ e ## Example 2: A particle with a charge of + 2.10 μC moves with a horizontal velocity of 4.00 m/s between two parallel plates, as shown below. The mass of the particle is 1.05 x 10^-2 g. The electric field between the plates has a magnitude of 25.0 N/C. The particle also experiences the effects of the gravitational field. What is the velocity of the particle the instant it strikes a plate? What plate does the particle strike? * q = + 2.10 x 10^-6 C * m = 1.05 x 10^-5 kg * E = 25.0 N/C * d = -1.00 m * *a* = -4.91 m/s² *F_net* = *F_e* + *F_g* *ma* = *qE* + *mg* *a* = *qE* + *mg* / *m* *a* = (2.10 x 10^-6 C)(25.0 N/C) + (1.05 x 10^-5 kg)(9.81 m/s²) / (1.05 x 10^-5 kg) *a* = -4.81 m/s² Particle hits the *lower plate* due to acceleration downwards. *V_ix* = 4.00 m/s *V_iy* = 0 m/s *V_fy* = √(*V_iy*)² + 2(*a*)(*d*) *V_fy* = √(0 m/s)² + 2(-4.81 m/s²)(-1.00 m) *V_fy* = -3.1016 m/s *V_f* = √(*V_fx*)² + (*V_fy*)² *V_f* = √(4.00 m/s)² + (-3.10 m/s)² *V_f* = 5.06 m/s θ = tan^-1 (3.10 m/s / 4.00 m/s) = 37.8° *V_f* = 5.06 m/s [37.8° Down from horizontal] # Millikan's Experiment - In 1897, JJ Thomson measured the charge to mass ratio of the electron. - Between 1906-1913, Millikan conducted the “Oil Drop Experiment” to find the charge on an electron. - Millikan placed charged oil droplets in a parallel plate capacitor and adjusted the potential difference so that the electric field exerted an upward force that *balanced* the downward gravitational force. ## Millikan's Assumptions - All *electrons* are identical *→* each has the same amount of charge. - Mass of an electron is so small that ± a few will *not* significantly change the mass of an oil droplet. - The amount of charge on the oil droplet will be a whole number multiple of the charge of one electron. ## Millikan's Findings - Millikan found charges that were all whole number multiples of -1.6 x 10^-19 C. - He concluded that charge is quantized. - The smallest or *elementary* charge is 1.6024 x 10^-19 C, the charge of 1 electron ## Example 3: An oil drop has a mass of 3.84 x 10^-15 kg. If it is suspended between two horizontal parallel plates where the electric field strength is 1.20 x 10⁴ N/C, what is the magnitude of the charge on the drop? * m = 3.84 x 10^-15 kg * E = 1.20 x 10⁴ N/C *F_e* = *F_g* *qE* = *mg* *q* = *mg* / *E* *q* = (3.84 x 10^-15 kg)(9.81 m/s²) / (1.20 x 10⁴ N/C) *q* = 3.14 x 10^-19 C ## Example 4: An oil drop (5.70 x 10^-16 kg) accelerates upward at a rate of 2.90 m/s² when placed between two horizontal parallel plates that are 3.50 cm apart. If the potential difference between the plates is 7.92 x 10² V, how many electrons are on the drop? * m = 5.70 x 10^-16 kg * a = + 2.90 m/s² * d = 0.0350 m * V = 7.92 x 10² V *F_net* = *F_e* + *F_g* *ma* = *qE* + *mg* *ma* = *q*(*V* / *d*) + *mg* *ma* - *mg* = *q*(*V* / *d*) *m*(*a* - *g*) = *q*(*V* / *d*) *q* = (*d* *m* *(*a* - *g*)) / *V* *q* = (0.0350 m)(5.70 x 10^-16 kg)(2.90 m/s² - (-9.81 m/s² )) / (7.92 x 10² V) *q* = 3.20 x 10^-19 C *e* = 3.20 x 10^-19 C / 1.60 x 10^-19 C = 2.00 *e* # Homework: - Part 1: Millikan's Experiment Practice - Part 2: Electric Energy, Power & Current Problems