course Phy 232
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
028. `Query 28
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Question: `qQuery introductory problems set 54 #'s 1-7.
Explain how to obtain the magnetic field due to a given current through a small current segment, and how the position of a point with respect to the segment affects the result.
Solution:
The specific value of the magnetic field is described using the equation:
B = k’*I*L/r^2*sin(theta)
I*L is the inverse square of the distance while the theta is the angle between IL and the radius.
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Question: `q`q** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is
B = k ' I L / r^2 * sin(theta). **
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Question: `qQuery principles and general college physics problem 17.34: How much charge flows from each terminal of 7.00 microF capacitor when connected to 12.0 volt battery?
I’m in university physics
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Question: `q`qCapacitance is stored charge per unit of voltage: C = Q / V. Thus the stored charge is Q = C * V, and the battery will have the effect of transferring charge of magnitude Q = C * V = 7.00 microF * 12.0 volts = 7.00 C / volt * 12.0 volts = 84.0 C of charge.
This would be accomplished the the flow of 84.0 C of positive charge from the positive terminal, or a flow of -84.0 C of charge from the negative terminal. Conventional batteries in conventional circuits transfer negative charges.
Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
** For current running in a circular loop:
Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located.
The sum of the fields from all the increments therefore has magnitude
B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is
B = k ' I / r^2 sum(`dL).
The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have
B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. **
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Question: `qQuery magnetic fields produced by electric currents.
What evidence do we have that electric currents produce magnetic fields?
Solution:
We have evidence that electric currents produce magnetic fields through the experiments that have been observed and preformed, like the coil having a current run through it and the metal balls being attracted to it. Also, if you think about how you are not supposed to place a magnet on or near your computer. The magnetic field of the magnetic disrupts the activity of the wires in your computer. There wouldn’t be an effect on the wires if they didn’t have some sort of magnetic field themselves.
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Question: `q`qSTUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field.
INSTRUCTOR COMMENT:
Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. **
How is the direction of an electric current related to the direction of the magnetic field that results?
** GOOD STUDENT RESPONSE:
The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. **
Query problem 17.35
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Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air?
Solution:
Dielectric constant of air = 1
Capacitor charge = Q
C = 0.2 F
d = 0.0022 m
Gauss’s Law:
E = 4*pi*k*Q/A
V = E*d = 4*pi*k*Q*d/A
C = Q/V = A/(4*pi*k*d)
A = C*4*pi*k*d
A = 0.2*4*pi*9*10^9*0.0022 = 4.98*10^7 m^2
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Question: `q`q** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation.
The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d).
Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C
If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have
A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m =
5 * 10^7 N m^2 / C^2 * C / ( J / C) * m =
5 * 10^7 N m^2 / (N m) * m =
5 * 10^7 m^2. **
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Question: `qQuery problem 17.50 charge Q on capacitor; separation halved as dielectric with const K inserted; by what factor does the energy storage change? Compare the new electric field with the old.
Note that the problem in the latest version of the text doubles rather than halves the separation. The solution for the halved separation, given here, should help you assess whether your solution was correct, and if not should help you construct a detailed and valid self-critique.
Solution:
C =Q/V
V = E*d; an electric field is independent of separation as long as the separation isn’t too big, I am going to assume that the separation can not be considered too big
V = Q/C
Q remains constant throughout because the capacitor is already charged to this particular charge
As the dielectric constant was inserted it increased the capacitance because it reduced the electric field. This in turn reduces the voltage by two factors one being 1/k and the other being ½.
C = Q/(1/2*1/k*V)
C = 2*k*Q/V
The capacitance is going to increase by 2k
Energy = ½*Q*^2/C
The energy will decrease by 2k
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Question: `q`q** For a capacitor we know the following:
Electric field is independent of separation, as long as we don't have some huge separation.
Voltage is work / unit charge to move from one plate to the other, which is force / unit charge * distance between plates, or electric field * distance. That is, V = E * d.
Capacitance is Q / V, ration of charge to voltage.
Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q).
The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation.
C will increase by factor k due to dielectric, and will also increase by factor 2 due to halving of the distance. This is because the electric field is independent of the distance between plates, so halving the distance will halve the voltage between the plates. Since C = Q / V, this halving of the denominator will double C.
Thus the capacitance increases by factor 2 k, which will decrease .5 Q^2 / C, the energy stored, by factor 2 k. **
STUDENT QUESTION:
I am very confused on the correct answer. I assumed the voltage would stay constant.
However, I think what the true answer is that voltage will increase by 1/k? My final answer is the same as yours, that the energy will increase by 2k.
INSTRUCTOR RESPONSE:
The capacitor is already charged, so Q remains constant.
The effect of the dielectric is to decrease the electric field, which by itself would decrease the voltage to 1/k of its former value, which increases the capacitance by factor k.
The effect of halving the distance is to decrease the voltage by another factor of 2, which increases the capacitance by factor 2.
Since Q remains constant, the energy .5 Q^2 / C decreases by factor 2 k.
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Self-critique (if necessary):
I did the overall equations right and the solving for the capacitance increasing by 2k. ??? I don’t understand how the energy decreases by 2k though??? If the equation for energy is ½*Q^2/C and C has increased by 2k wouldn’t the energy then decrease by 2k.
The energy does decrease by factor 2 k, as you say. I believe the given solution (to the extent we can locate it with the erroneous formating) says the same, as does the instructor response to the student question.
Self-critique Rating: 2
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Question: `qquery univ 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery.
What is the capacitance of this capacitor?
Solution:
A = 0.16 m^2
d = 0.0047 m
V = 12 V
k =9*10^9 N*m^2/C^2
Electric field:
E = V/d = 12 V/0.0047 m = 2553.19 V/m
The electric field is created by two plates with equal and opposite charges; therefore:
4*pi*k*sigma = 2553.19 V/m
Sigma = 2553.19 V/m/(4*pi*9*10^9 N*m^2/C^2) = 2.26*10^-8 C/m^2
Q = A*sigma = 0.0256 m^2*2.26*10^-8 C/m^2 = 5.78*10^-8 C
C = Q/V = 5.78*10^-8 C/12 V = 4.82*10^-11 F
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Question: `q`q** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows:
If the 4.7 mm separation experiences a 12 V potential difference then the electric field is
E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx.
Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have
4 pi k sigma = 2250 V / m and
sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2.
The area of the plate is .0256 m^2 so the charge on a plate is
.0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C.
The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads.
The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor.
The work to move a charge Q across an average potential difference Vave is Vave * Q.
Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is
energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C.
Since the unit V / m * C is the same as J / C * C = J, we see that the energy is
3.4 * 10^-9 J.
Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy.
Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C).
This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C)
Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. **
query univ 24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery.
If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored?
The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies
electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx..
The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have
Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx..
The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J.
If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C.
The energy stored will also be halved, since V remains the same but Q is halved.
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Question: `qquery univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q.
What is the electric-field energy density at distance r < R from the center of the sphere?
What is the electric-field energy density at distance r > R from the center of the sphere?
Solution:
Integrate the energy stored in the volume of the sphere from r to infinity.
Energy density = C*V^2/(Vol*2) = C*V^2/(d*A*2)
C = epsilon0*A/d
V = E*d
Energy density = episilon0*E^2/2
For the charged sphere of r>R
E = Q / (4*pi*epsilon0*r^2)
Energy density = Q^2/(32*pi^2*episilon0*r^4)
Integrate the energy density from r = R to infinity:
Q^2/(8*pi*epsilon0*R) = k*Q^2/(2*R)
W = k*Q/R dQ
Integrate the work from q = 0 to q = Q
And when k = 1/(4*pi*epsilion0)
W = k*Q^2 / (8*pi*epsilon0*R)
The work and the energy are equal
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Question: `q`q** The idea is that we have to integrate the energy density over all space. We'll do this by finding the total energy in a thin spherical shell of radius r and thickness `dr, using this result to obtain an expression we integrate from R to infinity, noting that the field of the conducting sphere is zero for r < R.
Then we can integrate to find the work required to assemble the charge on the surface of the sphere and we'll find that the two results are equal.
Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is
Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates.
Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that
Energy density = .5 epsilon0 E^2, or in terms of k
Energy density = 1 / (8 pi k) E^2,
Since your text uses epsilon0 I'll do the same on this problem, where the epsilon0 notation makes a good deal of sense:
For the charged sphere we have for r > R
E = Q / (4 pi epsilon0 r^2), and therefore
energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4).
The energy density between r and r + `dr is nearly constant if `dr is small, with energy density approximately Q^2 / (32 pi^2 epsilon0 r^4).
The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr.
The expression for the energy lying between distance r and r + `dr is therefore approximately energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr.
This leads to a Riemann sum over radius r; as we let `dr approach zero we approach an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r.
To get the energy between two radii we therefore integrate this expression between those two radii.
If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere.
This integral gives us Q^2 / (8 pi epsilon0 R), which is the same as k Q^2 / (2 R).
The work required to bring a charge `dq from infinity to a sphere containing charge q is k q / R `dq, leading to the integral of k q / R with respect to q. If we integrate from q = 0 to q = Q we get the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get k Q^2 / (2 R).
Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = k Q^2 / (8 pi epsilon0 R).
So the energy in the field is equal to the work required to assemble the charge distribution. **
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&#Good responses. See my notes and let me know if you have questions. &#