1. Voltage due to point sources
Contents:
Voltage due to one point charge
Whiteboards
Voltage due to many point charges
Example
Cute Voltage problems

2. kq1q2 .dr Voltage due to Point Sources r Definition: V = W q
W = Fs, but what work to bring q2 from infinity to r? r q1 q2
The work is:
kq1q2 r
It’s simple, just the definite integral of
kq1q2 .dr r2
From infinity to r

3. What is your gravitational potential on the surface of the earth?
r = 6.38x106 m,
m = 5.97x1024 kg
A van de Graaff generator has an 18 cm radius dome, and a charge of μC. What is the voltage at the surface of the dome?
(41 kV) sheet of rubber!!
(-6.24x107 J/kg) – 1 kg mass, KE, escape, zero energy, big bang paradox

4. Voltage due to point charge
Whiteboards:
Voltage due to point charge
1 | 2 | 3

5. Lauren Order is 3. 45 m from a -150. C charge
Lauren Order is 3.45 m from a C charge. What is the voltage at this point?
V = kq/r, r = 3.45 m, q = -150x10-6 C
V = -3.91x105 V
-3.91x105 V

6. Gerry Mann-Derr is in orbit 3.40x106 m above the surface of the earth.
Me = 5.97x1024 kg, re = 6.38x106 m
solution
-4.07x107 J/kg

7. Rhoda Tiller has a gravitational potential of -1
Rhoda Tiller has a gravitational potential of -1.90x106 J/kg near the moon. What is her distance from the center of the moon if it has a mass of 7.35x1022 kg? What is her height above the surface if the moon has a radius of 1.737x106 m?
solution
2.58x106 m, 8.43x105 m above the surface, or 843 km

8. Alex Tudance measures a voltage of 25,000 volts near a Van de Graaff generator whose dome is 7.8 cm in radius. What is the charge on the dome?
V = kq/r, r = .078 m, V = 25,000 V
q = 2.17x10-7 C = .22 C
0.22 C

9. Ashley Knott reads a voltage of 10,000. volts at what distance from a 1.00 C charge?
V = kq/r, V = 10,000 V, q = 1.00x10-6 C
r = .899 m
0.899 m

10. Voltage is not a vector!!!!!! Voltages in non linear arrays
Find the voltage at point A: A
+1.5 C
+3.1 C
75 cm
190 cm Q1 Q2
Voltage is not a vector!!!!!!

13. Find the voltage at point A:
+1.5 C
+3.1 C
75 cm
190 cm Q1 Q2
Voltage at A is scalar sum of V1 and V2:
Voltage due to Q1:
V1 = kq1 = k(1.5x10-6) = x104 V
r ( )
Voltage due to Q2:
V2 = kq2 = k(3.1x10-6) = x104 V
r ( ) +
2.6x104 V
And The Sum Is…

14. Whiteboards:
Voltage charge arrays
1 | 2 | 3

15. Find the voltage at point B
+1.1 C
+2.1 C B
91 cm
138 cm Q1 Q2
V1 =
V2 =
V1 + V1 = = 25,000 V
25,000 V

16. Find the voltage at point C
38 cm
85 cm Q1 Q2
+1.1 C
V1 =
V2 =
V1 + V1 = = -14,000 V
-14,000 V

17. = k(-14.7E-6)/√(12+42) + k(17.2E-6)/(22+22) = 22617.44323
Each grid is a meter. If charge A is μC, and charge B is μC, calculate the voltage at the origin: y A B
V = kq/r + kq/r
= k(-14.7E-6)/√(12+42) + k(17.2E-6)/(22+22) =
+2.26E4 V x

18. Find the voltage in the center of a square 45
Find the voltage in the center of a square 45.0 cm on a side whose corners are occupied by 12.0 C charges.
The center is ( )/2 from all of the charges
one charge’s V = V
All four: V = 1.36x106 V
1.36x106 V