Download presentation

Presentation is loading. Please wait.

Published byHorace Dennis Modified over 2 years ago

2
Voltage due to point sources Contents: Voltage due to one point charge Whiteboards Voltage due to many point charges Example Whiteboards Cute Voltage problems Example Whiteboards

3
Voltage due to Point Sources TOC Definition: V = ΔE p q ΔE p = W = Fs, but what work to bring q 2 from infinity to r? q2q2 q1q1 r It’s simple, just the definite integral of kq 1 q 2.dr r 2 From infinity to r The work is: kq 1 q 2 r

4
Voltage due to Point Sources TOC Definition: V = ΔE p q ΔE p = W = Fs, q1q1 r The voltage at point A is: V = kq r V = voltage at distance r q = charge of q 1 r = distance from q 1 A A van de Graaff generator has an 18 cm radius dome, and a charge of 0.83 μC. What is the voltage at the surface of the dome? V = kq/r = (8.99E9)(0.83E-6)/(0.18) = 41453.88889 ≈ 41 kV 10,000 V/in, +q and –q, sheet of rubber

5
Whiteboards: Voltage due to point charge 11 | 2 | 323 TOC

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

7
.22 C W 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

8
.899 m W 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

9
Voltages in non linear arrays TOC Q2Q2 Q1Q1 +1.5 C +3.1 C 190 cm75 cm Find the voltage at point A: A Voltage is not a vector!!!!!!

10
TOC Q2Q2 Q1Q1 +1.5 C +3.1 C 190 cm75 cm Find the voltage at point A: A Voltage at A is scalar sum of V 1 and V 2 : Voltage due to Q 1 : V 1 = kq 1 =k(1.5x10 -6 )= 1.27x10 4 V r (.75 2 +.75 2 ) Voltage due to Q 2 : V 2 = kq 2 =k(3.1x10 -6 )= 1.36x10 4 V r (.75 2 +1.9 2 ) + 2.6x10 4 V And The Sum Is…

11
Whiteboards: Voltage charge arrays 11 | 2 | 323 TOC

12
25,000 V W V 1 =10867.03297 V 2 =13680.43478 V 1 + V 1 =24547.46775 = 25,000 V Find the voltage at point B Q2Q2 Q1Q1 +1.1 C +2.1 C 138 cm91 cm B

13
-14,000 V W V 1 =-39587.58847 V 2 =+26023.68421 V 1 + V 1 =-13563.90426 = -14,000 V Find the voltage at point C Q2Q2 Q1Q1 -4.1 C +1.1 C 38 cm 85 cm C

14
A B Each grid is a meter. If charge A is -14.7 μC, and charge B is +17.2 μC, calculate the voltage at the origin: y x V = kq/r + kq/r = k(-14.7E-6)/√(1 2 +4 2 ) + k(17.2E-6)/(2 2 +2 2 ) = 22617.44323 +2.26E4 V

15
1.36x10 6 V W The center is (.45 2 +.45 2 )/2 from all of the charges one charge’s V = 339034.1314 V All four:1356136.525 V = 1.36x10 6 V Find the voltage in the center of a square 45.0 cm on a side whose corners are occupied by 12.0 C charges.

Similar presentations

OK

Consider a point charge, +q fixed at the origin A positive test charge,q 0 is placed at A, a distance r A Coulomb’s law determines the magnitude of repulsive.

Consider a point charge, +q fixed at the origin A positive test charge,q 0 is placed at A, a distance r A Coulomb’s law determines the magnitude of repulsive.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on plant layout design Ppt on do's and don'ts of group discussion topic Download ppt on surface area and volume for class 9th Ppt on road accidents in the philippines Ppt on online banking project Ppt on various properties of air Ppt on south african culture tattoos Ppt on accounts payable process Ppt on structure of atom class 11th Ppt on dda line drawing algorithm