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course PHY 232

4/4/2011; 16:00;I only have a question about last problem, though I have probably made some dumb errors in the rest of my work.

110323 Physics II

`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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8 * 10^5 Newtons / Coulomb * 2 * 10^-6 C * (10cm - 30cm) =

= - .2 m * 1.6N = - .32J

@& Good, but it takes positive energy to move that charge:

To move the charge at a constant velocity toward the origin takes a force equal and opposite to the force exerted by the field, so the necessary force is in the direction of motion and the work will be positive.

The work is just `dW = ave force * displacement, so

`dW = -1.6 N * (10 cm - 30 cm) = 32 N * cm = 32 N * (.01 m) = .32 Joules.

*@

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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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The distance between the center of the first charge and the moment of charge on the second charge = sqrt(30^2 + 40^2) cm = 50 cm = r

k = 9*10^9

q1 = charge 1 = 3*10^-6 C

q2 = charge 2 = 1*10^-6 C

r = .5m

F = k*q1*q2/r^2 = (9*10^9 N*m^2 /C^2) * (3*10^-6 C) * (1*10^-6 C)/(.25m^2)

=9*4 *3*10^-3 N = 36*3*10^-3 N = 108 milli-N = .108 N

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What is the direction of this force?

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They are both positive => repulsion, away from the other charge.

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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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for a small distance the force is near concstant = .108 N * 1 cm = .108 * .01J = 1.08 milli - 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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there are 4 micro C involved so… 1.08 *10^-3 J / (4*10^-6 C) = about .252 * 10 ^ 3 J/C

=252 Joules / Coulumb

@& Only one microCoulomb of charge is being moved.*@

@&

The work required to move the charge 1 cm closer would be

F = .108 N * .01 m = .00108 Joules.

Dividing this work by the 1 microCoulomb of charge being moved we get

`dW / q = .00108 J / (10^-6 C) = 1080 J / C.*@

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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, E = F / q1 = k * q1 * q2 /(d^2 *q2)

= k * q1 / (d^2) = 9^10^9 [N*m^2/C^2]* 3*10^-6 C/.5^2m^2

=27*10^3 [N*m^2 / C]/.25m^2 = 108 kN/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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Volt = J/C

Volt = 1.5

ampere = .2

J/min = ?

Volt = Joules / Coulomb

Ampere = Coulumb/s

J/s = Volt * Amp = 1.5 J/C * .2C/s = 0.3 J/s = 18 J/min

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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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Volts = 6 J/C

Apmere = .050 C/s

t = .2 s

6 J/C * .050 C/s * .2 s = .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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the lightbulb is 250 milli - Amps = .25 C/s

=> 3.75 J

.25 * 1.5*10 J = 4 J

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`q004. A magnet spins at 10 revolutions per second in the vicinity of a coil of wire enclosing a total area of 1000 cm^2. The magnet creates a maximum field of .04 Tesla within that area. At what average rate is the magnetic flux through the coil changing?

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Area = .1m^2

T = .04

a Tesla = kg / (C*s)

10 revolutions / second.

=> omega = 10 *2pi = 20 pi radian’s / sec

The modeling of the cycle is cos (20pi*t) radians

T*Area = J/s

.04 T / .1m^2 = .004 T/m^2 = .004 J*rad/s/m^4

however the field is reversed each half of a time so the magnetic flux can be modeled

by 2* .4 J/s = 8milli Joules/sec/m^4

magnetic flux = .8 J/s/m^4 * cos(20Pi*t)radians

Phi(t) = . 8* cos(20Pi*t)J/s/m^4

the change in the

@&

The maximum magnetic flux is

phi_B = 1000 cm^2 * .04 T = .004 T * m^2.

The minimum flux is -.004 T * m^2.

The magnitude of flux change between max and min, or between min and max, is therefore .008 T * m^2.

This magnitude of flux change occurs twice per revolution, therefore 20 times per second. The corresponding time interval is .05 second.

So the average rate of flux change is approximately

.008 T * m^2 / (.05 second) = .160 T m^2 / sec = .16 volt.

The true average rate is found by integrating the instantaneous rate with respect to clock time, and differs slightly from this value.

*@

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`q005. Charge Q creates electrostatic flux 4 pi k Q, with k = 9 * 10^9 N m^2 / Coulomb^2. A 5 meter length of wire contains a uniformly distributed charge of 8 microCoulombs.

How much charge does a 25 cm length of the wire contain?

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25/500 = .05 cm/cm * 8 *10 ^-6 C = .4 * 10^-6 C

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What will therefore be the electrostatic flux through a piece of pipe 25 cm long, which surrounds the wire?

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Electrostatic Flux = 4 pi k Q

= 4Pi * 9 * 10^9 (N m^2 / Coulomb^2) * (0.4 * 10^-6) C = 14.4 Pi * 10^3 N*m^2 / C

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If the diameter of the pipe is 5 cm, then how much flux is there per square centimeter of the pipe?

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(14.4 Pi * 10^3 N*m^2 / C) / (.05^2 m^2)

(14.4 Pi * 10^3 N*m^2 / C)/ (.0025 m^2)

= ( 57.6) Pi * 10^5 N/C

= 5.76 Pi * 10^6 N/C

= 5.76 Pi Mega-N/C

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How much is that per square meter?

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1/1cm^2 * 10000 cm^2/m^2 = 1 * 10^4 / m^2

there are 1 * 10^4 cm^2 per m^2

=> There are 56.7 * Pi Giga-N/C

@& Good, but you ended up dividing by the area twice. That's compounded by the fact that you didn't get the right area, but your procedure is very good, and both errors are easy to correct.

5 meters of the wire contains 8 microCoulombs, so 25 cm of the wire contains .4 microCoulombs. This can be reasoned as follows:

.25 cm is 1/20 of 5 meters, so the charge enclosed is 1/20 of 8 microCoulombs, or .4 microCoulombs.

Alternatively, 8 microCoulombs on 5 meters of wires implies charge density 8 microCoulombs / (5 meters) = 1.6 microCoulombs / meter. So a .25 meter section has charge 1.6 microCoulombs / meter * .25 meter = .4 microCoulombs = 4 * 10^-7 C.

The electrostatic flux through the pipe is therefore 4 pi k Q = 4 pi * 9 * 10^9 N m^2 / C^2 * 4 * 10^-7 C = 14 400 pi N m^2 / C.

The flux exits radially through the circular surface of the pipe, so the area through which it exits is (circumference * length) = (2 pi * 5 cm) * (25 cm) = 250 pi cm^2 = .0025 pi m^2.

The flux / area is therefore

phi_E / area = (14 400 pi N m^2 / C) / (.0025 pi m^2) = 5 800 000 N / C,

approximately.

*@

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`q006. Suppose the blackboard is uniformly covered by electrostatic charge, with a total charge of 60 microCoulombs. We place a piece of tile 1 cm from the board, parallel to and near the center of the board. The tile dimensions are 15 cm by 15 cm. The tile does not affect the electric field, which effectively passes right through it.

What must be the direction of the electric field relative to the surface of the tile?

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perpendicular

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What do you think is the electrostatic flux through the tile?

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assuming 10 tiles cover the board vertically and 25 horizontally => the charge on of the board = 8 micro-Coulombs * 1/250 = 32 nano-Coulombs

The charge is applied to both sides of the board => the charge effecting the tile is on half that exerted by the board = 16 nano-Coulombs on the tile.

The electrostatic field on the tile is F/q_board = 9*10^9 N*m^2/C^2*16*10^-9C/1cm^2

=(144 N*m^2/C) /cm^2

@& This isn't a point charge so Coulomb's Law isn't appropriate to this situation. To use Coulomb's Law you would have to integrate (charge density * dA) * Coulomb force from region dA, over the entire area of the board. The Coulomb force would be a vector in space. As it turns out all the components except that perpendicular to the board would pretty much cancel out. We'll probably look at this integral in class--it's really not that bad.

However since we have the symmetries we do, we want to use the symmetry argument and the flux picture.*@

@& Here's the argument, with a little hand-waving; we'll fill in the details later.

By symmetry the direction of the field is perpendicular to the board.

The flux through the tile comes from an area of the board equal to that of the tile. The flux through the tile depends on the charge within that area of the board.

The area of the tile is 15 cm * 15 cm = 225 cm^2 = .0225 m^2.

The board has an approximate area of 3 m^2 and contains 60 microCoulombs of charge. Your estimate of the area might differ from this, but this estimate would imply that the charge density on the board is 60 microCoulombs / (3 m^2) = 20 microCoulombs / m^2. Multiplying this by .0225 m^2 would imply a charge of about .25 microCoulombs within the 15 cm x 15 cm square.

Alternatively we might estimate that to cover the board would take about 10 rows of 20 tiles, or 200 tiles. So the charge corresponding to the area of one tile would be 60 microCoulombs / 200 = .30 microCoulombs.

The estimates are pretty consistent. We'll use the .30 microCoulombs as a basis for our subsequent results.

The flux corresponding to .30 microCoulombs of charge is

phi_E = 4 pi k Q = ... = 10 800 pi N / C^2.

Now the flux from that region extends out of the front of the board, as well as out the back. You can imagine another tile held just behind the board, so that the flux exits through the total area of two tiles.

It follows that

flux / area = 10 800 pi N / C^2 / (.045 m^2) = 7 000 000 N / C, approximately.

With the tile close to the board, most of the board is much further away from the tile than the closest part of the board, and the far-away part of the board has no significant effect on the electric field through the tile. This allows us to make our symmetry argument and conclude that the field is perpendicular to the board and the tile. If we move the tile far away from the board, though, no part of the board is that much further away from the tile than the closest part, so we have to consider the entire board, and our symmetry argument breaks down. We can no longer assume that the field passes through the tile at a right angle. This is a bit of a hand-waving argument and you might not find it completely satisfactory, but accept it for the moment.

If we move the tile tile 2 cm further from the board, it's still much, much closer to the nearest point of the board than the furthest, and our symmetry argument still holds. So there will be no change in our result.

Our conclusion: As long as we remain close enough to the board, the electric field we experience will not depend on just how close we are.*@

1 cm^2 = 10000^-1 m^2

(144 N*m^2/C) /(225 m^2) = 1.44 Mega-N/C

The electrostatic flux = F/q * Area= (1.44 Mega-N/C) * 225 cm^2

= 324 * 10^6 N*cm^2/C

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What is the electrostatic flux per square centimeter, and per square meter of the tile surface?

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(324 * 10^6 N*cm^2/C) / 225 cm^2 = 1.44Mega-N/C

(324 * 10^6 N*cm^2/C) / .0225 m^2 = 14.4 Giga-N/C

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How much do you think this result will change if we move the tile 2 cm further from the board?

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By the inverse of the square of the distance => if the distance is doubled the applied force/coulomb will be fourthed or fourpled^-1.

=> a force / coulomb of 3.6 * 10^9 N/C

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????????????Question number 007 is rather foggy to me I am not sure if I got any part of it correct due to not having a intuitive sense of units etc. Please do let me know what I am doing right if any thing and how to understand what is going on.

It seemed to me that this question was relevant to the coil around the spool in class where you attached a battery to the coil and placed a bolt in the center of the spool. If this is right or wrong as to what is going on I would like to know that as well!

Thank you so much for your help in this matter!

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`q007. A current of .25 amps is driven by a 1.5 volt battery through a 100-turn coil of radius 2 cm.

What magnetic field results at the center of the coil?

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Apmhere = .25 C/s

Volts = 1.5 J/C

r = 2 cm = .02 m

The cross sectional area = 4Pi cm^2 =about 12.6 cm^2

the potential energy of the coil is .375J/s

=.375 N*m/s [kg*m^2/s^3]

electric field = k * .25C/s / 4cm^2 = (9*10^9 N*m^2/C^2)*(.25C/s) / (.0004m^2)

= (2.25/4) * (10^9*10^4) (N/C*s) = 5.63 * 10^12 (N/(C*s));

which means that the field is reinforced each second for the field.

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What therefore is the magnetic flux within the coil?

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CSA = 12.6 cm^2

E = 5.63 * 10^12 (N/C*s)

E*CSA = .63 * 10^12 (N/C*s) *12.6 cm^2

=7.94 *10^12 (N*cm^2) / (C*s)

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If the battery is removed the current in the coil will quickly fall to zero. If this occurs in 20 milliseconds, then at what average rate must the magnetic flux change?

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the change in time is .02 sec, the change in the flux is its finial strength - the original strength = - original strength.

the average change with respect to time is - original strength / .02 sec = - original strength * 50 /sec

==7.94 *10^12 (N*cm^2) / (C*s) * 50 / sec = about 4.97 * 10^14 N*cm^2) / (C*s^2)

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"

@& You are on the wrong track here, as you suspected. The current gives rise to a magnetic field. There's no electric field present inside the coil.

The magnetic field of a small segment wire of length `dL will be `dB = k ' * I `dL / r^2. By the right-hand rule the field will be perpendicular to the plane of the loop.

If we assume that the coil is pretty much circular (as opposed to cylindrical) the contributions of all segments reinforce, and the field is

B = k ' * I L / r^2,

where L is the total length of the wire.

Every 2 cm loop of wire has circumference 4 pi cm. There are 100 loops, so their total length is 400 pi cm.

Thus

B = 10^-7 T * m / amp * .25 amps * 400 pi cm / (2 cm)^2

= 10^-7 T * m / amp * .25 amps * 4 pi meter / (.02 m)^2

= 8 * 10^-4 T,

very approximately.

Note that the magnetic field of the Earth is on the order of 10^-4 Tesla. So this field would pretty seriously mess with a compass.

*@