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course Phy 201
4/4 8 3.23.11
`q001. How much energy does it take to move a 2 microCoulomb charge from x = 30 cm to x = 10 cm, given a field of 8 * 10^5 Newtons / Coulomb directed in the positive x direction?
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8x10^5 N/C * 2x 10^-6 C=1.6 N
`dW=+1.6 N * (10 cm-30 cm)=-32 N/cm=-.32 J
.32 J/2x10^-6 C=1.6 x 10^5 J/C
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`q002. How much force is experienced by a 1 microCoulomb charge at the point (30 cm, 40 cm) by a 3 microCoulomb charge at the origin?
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.108 N
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What is the direction of this force?
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Away from origin, extending from the last point.
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How much work would it take to move the 1 microCoulomb charge one centimeter closer to the origin?
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.108 N * 1 cm=.00108 J
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How many Joules does it take per Coulomb of moving charge to move that one centimeter?
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`dPE/ q= .00108 J/ 1x10^-6 C=1080 J/C (volt)
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What is the electric field at the point (30 cm, 40 cm) due to a 3 microCoulomb charge at the origin?
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The electric field at a point where a 1 microCoulomb charge experiences a force of .108 N is
E = .108 N / (10^-6 C) = 1.08 * 10^5 N / C.
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`q003. A volt is a Joule of potential energy per Coulomb.
How much does chemical potential energy stored in a 1.5 volt battery change, per minute, as it drives a .20 amp current?
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.20 amps=.2 C/s
1.5 volt=1.5 J/C
.2 C/s*1.5 J/C=.3J/s
.3J/s * 60 s=18 J
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By how much does the electrostatic potential energy of a capacitor change in 200 milliseconds, if the capacitor is at 6 volts and drives a current of 50 milliamps?
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6 volts=6 J/C 50mA=.050 C/s
6 J/C * .050 C/s=.3 J/s
In 200 mseconds, the energy loss is .3 J/s *.2 s=.06 J
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How much energy does it take to light your lightbulb at 1.5 volts for 10 seconds?
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I’m afraid I’m over simplifying this:
1.5 volts= 1.5 J/C over 10 seconds
1.5 J/C * 10 seconds==I know that’s not right because I don’t end with a unit of energy.
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A current of .20 amps is .20 Coulombs / second. If the current lasts a minute, then the charge passing through the current in that time is .20 C / s * 60 s = 12 C.
1.5 volts means 1.5 Joules / Coulomb, so the loss of potential energy is
1.5 J / C * 12 C = 18 J.
A capacitor driving a current of 50 milliamps = .05 amps .05 C / s, for 200 milliseconds = .2 seconds, has charge .05 C / s * .2 s = .01 C passing through it in that time interval.
.01 C passing through a 6 volt increase in voltage would increase its PE by .01 C * 6 J / C = .06 J.
If for example the lightbulb you observed in class had a current of 100 milliamps at voltage 3 volts, then in 10 seconds the charge flowing through it would be 100 milliAmps * 3 volts = .1 C / s * 3 J / C = .3 Joules.
In symbols, these calculations are summarized by
`dQ = I * `dt and
| `dPE | = | `dQ * V |.
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