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

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Presentation on theme: "Voltage due to point sources Contents: Voltage due to one point charge Whiteboards Voltage due to many point charges Example Whiteboards Cute Voltage."— Presentation transcript:

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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) = ≈ 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  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 Q1Q  C +3.1  C 190 cm75 cm Find the voltage at point A: A Voltage is not a vector!!!!!!

10 TOC Q2Q2 Q1Q  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  ( ) Voltage due to Q 2 : V 2 = kq 2 =k(3.1x10 -6 )= 1.36x10 4 V r  ( ) + 2.6x10 4 V And The Sum Is…

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

12 25,000 V W V 1 = V 2 = V 1 + V 1 = = 25,000 V Find the voltage at point B Q2Q2 Q1Q  C +2.1  C 138 cm91 cm B

13 -14,000 V W V 1 = V 2 = V 1 + V 1 = = -14,000 V Find the voltage at point C Q2Q2 Q1Q  C +1.1  C 38 cm 85 cm C

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

15 1.36x10 6 V W The center is  ( )/2 from all of the charges one charge’s V = V All four: 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.


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