# 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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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

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

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

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

-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?

.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

.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

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!!!!!!

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…

Whiteboards: Voltage charge arrays 11 | 2 | 323 TOC

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

-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

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

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.

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