1. Millikan Experiment Worksheet- Answer Key (Front & Back) Fe E Fg

2. Front Side
1. Fg = 1.9x10-15 N [D] “stationary” means Fe= -Fg so Fe = 1.9x10-15 N [U]
E = 6.0x103 N/C (if top plate is “-” then [U])
a. q = Fe /E
= 1.9x10-15 N [U]
6.0x103 N/C [U]
=+ 3.2x10-19 C
b. If the top plate is “-”
then only a “+” charge
would be attracted
#e lost = 3.2x10-19 C
1.6x10-19 C
= 2 electrons lost Fe + E Fg

3. 2. m = 6.5x10-14 kg
E = 4.0x106 N/C (top plate must be “-” to suspend a “+” drop)
a. “suspended” means Fe= -Fg so Fe = 1.9x10-15 N [U]
q = Fe /E
= 1.9x10-15 N [U]
6.0x103 N/C [U]
=+ 3.2x10-19 C
b. If the top plate is “-”
then only a “+” charge
would be attracted
#e lost = 3.2x10-19 C
1.6x10-19 C
= 2 electrons lost Fe + E Fg

4. E + Fe Fg Has 4 protons so q=(+4)(1.6x10-19C) = +6.4x10-19C
(top plate must be “-” to suspend a “+” drop)
Fg = 6.4x10-13 N [D]
“suspended” means Fe= -Fg so Fe = 6.4x10-13 N [U]
E = ?
E = Fe /q
= 6.4x10-13 N [U]
+6.4x10-19C
= 1.0x106 N/C [U] Fe E + Fg

5. + E Fe Fg a a= 18.4m/s2 [D] m = 4.95x10-15 kg
a. Fe= ? a. a> 9.8 m/s2 means Fe is down
Fnet= ma
= (4.95x10-15 kg)(18.4 m/s2 [D])
= 9.11x10-14 N [D]
Fg = mg
= (4.95x10-15 kg)(9.8 m/s2 [D])
= 4.85x10-14 N [D]
so Fe = Fnet – Fg
= 9.11x10-14 N [D] x10-14 N [U]
= 4.26x10-14 N [D]
b. If top plate is “+”, what kind of charge is it?
Positive charge would be repelled by a positive plate/attracted by a
negative plate. + Fe Fg a E

6. + E a Fe Fg 5. a= 12.4m/s2 [D] m = 5.95x10-15 kg = 1.55x10-14 N [D]
b. If top plate is “+” what kind of charge is it? - Positive charge again
a. a> 9.8 m/s2 means Fe is down
Fnet= ma
= (5.95x10-15 kg)(12.4 m/s2 [D])
= 7.38x10-14 N [D]
Fg = mg
= (5.95x10-15 kg)(9.8 m/s2 [D])
= 5.83x10-14 N [D]
Fe = Fnet - Fg
= 7.38x10-14 N [D] x10-14 N [D]
= 1.55x10-14 N [D] + a E Fe Fg

7. E Fe q a Fg 6. a= 8.0m/s2 [D] m = 1.5x10-25 kg
I have randomly assigned plates here; it could easily have been the other way
a< 9.8m/s2 means Fe is up
Fnet= ma
= (1.5x10-15 kg)(8.0 m/s2 [D])
= 1.2x10-14 N [D]
Fg = mg
= (1.5x10-15 kg)(9.8 m/s2 [D])
= 1.5x10-14 N [D]
Fe = Fnet - Fg
= 1.2x10-14 N [D] – 1.5x10-14 N [D]
= - 3x10-15 N [D] or + 3x10-15 N [U] Fe q a E Fg

8. E + Fe Fg Back Side 1.a) m = 8.22x10-11 kg Fg = mg q= Fe /E
E = 4.36x107 N/C
for a positively charged oil drop to be “balanced” means
Fe= -Fg and that the top plate must be negative
Fg = mg
= (8.22x10-11 kg)(9.8 m/s2 [D])
= 8.06x10-10 N [D]
Fe = - Fg
= 8.06x10-10 N [U]
q= Fe /E
4.36x107 N/C [U]
= x10-17 C Fe E + Fg

9. E + Fe a Fg Back Side 1.b) same oildrop so
m = 8.22x10-11 kg, Fg = 8.06x10-10 N [D]
but a=6.1m/s2 [D]
Fe= ? Fnet = ma
= (8.22x10-11 kg)(6.1 m/s2 [D])
= 5.01x10-10 N [D]
Fe = Fnet - Fg
= 5.01x10-10 N [D] x10-10 N [D]
= -3.05x10-10 N [D]
OR x10-10 N [U] Fe a E + Fg

10. + E Fg Fe a Back Side 1.c) same oildrop so
m = 8.22x10-11 kg, Fg = 8.06x10-10 N [D]
but a=17.4m/s2 [D]
Fe= ? Fnet = ma
= (8.22x10-11 kg)(17.4 m/s2 [D])
= 1.43x10-9 N [D]
Fe = Fnet - Fg
= 1.43x10-9 N [D] x10-10 N [D]
= 6.2x10-10 N [D] + E Fg Fe a

11. + E Fe Fg “balanced” therefore Fe = - Fg E= ? q=(+16)(1.6x10-19C)
Back Side
2.a) m = 4.17x10-9 kg and an excess of 16 protons (lost 16 electrons)
“balanced” therefore Fe = - Fg
E= ? q=(+16)(1.6x10-19C)
q = +2.56x10-18C
Fg = mg
= (4.17x10-9 kg)(9.8 m/s2 [D])
= 4.09x10-8 N [D]
Fe = 4.09x10-8 N [U]
E = Fe /q
= 4.09x10-8 N [U]
+2.56x10-18C
= 1.60x1010 N/C Fe + E Fg

12. + E Fe Fg but a=42.5m/s2 [D] = (4.17x10-9 kg)(42.5 m/s2 [D])
Back Side
2.b) same drop so m = 4.17x10-9 kg and Fg = 4.09x10-8 N [D]
but a=42.5m/s2 [D]
Fe = ?
Fnet = ma
= (4.17x10-9 kg)(42.5 m/s2 [D])
= 1.77x10-7 N [D]
Fe = Fnet - Fg
= 1.77x10-7 N [D] x10-8 N [D] = 1.36x10-7 N [D] + E Fe Fg

13. + E Fe a Fg but a=0.279m/s2 [D] = (4.17x10-9 kg)(0.279 m/s2 [D])
Back Side
2.c) same drop so m = 4.17x10-9 kg and Fg = 4.09x10-8 N [D]
but a=0.279m/s2 [D]
E = ?
Fnet = ma
= (4.17x10-9 kg)(0.279 m/s2 [D])
= 1.16x10-9 N [D]
Fe = Fnet - Fg
= 1.16x10-9 N [D] x10-8 N [D] = x10-8 N [D]
OR x10-8 N [U]
E = Fe /q
= 3.97x10-8 N [U] Fe + a E Fg
+2.56x10-18C
= 1.55x10-10 N/C [U]