course Phy 242 bwlassignment #021
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20:26:44 Query introductory set #1, 9-16 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
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RESPONSE --> Using the equation: E = kq/r^2 we can solve for the uniform electric field (E). The k is given and the q (charge) and r (radius) are known. Then we take the uniform electric field (E) and sove for the potential difference given the distance between the two points. V = E*'ds
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20:26:55 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
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RESPONSE --> understood
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20:30:16 Explain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.
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RESPONSE --> We can take the equation: F = q*'dV/'ds and solve for 'dV 'dV = F'ds/q
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20:30:37 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
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RESPONSE --> understood
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20:31:07 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
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RESPONSE --> We can use the equation: 'dW = q*E*'ds
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20:31:44 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
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RESPONSE --> I understand I should have actually explained the answer.
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20:32:00 ** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
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RESPONSE --> ok
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20:37:26 In your own words explain the meaning of the electric field.
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RESPONSE --> An electric field is a pattern of lines of force that surround an electric charge.
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20:37:36 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
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RESPONSE --> ok
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20:41:52 In your own words explain the meaning of voltage.
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RESPONSE --> Voltage is the difference in electrical potential between two points.
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20:41:59 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
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RESPONSE --> ok
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z喳ɂ^DZПغ assignment #022 { Physics II 12-12-2006
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20:47:35 Query problem set 1 #'s 17-24 If we know the initial KE of a particle, its charge and the uniform electric field in which it moves, then if the net force on the particle is due only to the electric field, how do we find the KE after the particle has moved through a given displacement?
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RESPONSE --> I am not sure what would be the best route to take to solve this problem.
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20:48:44 ** GOOD STUDENT SOLUTION: Given KE0, q, E, `ds: First we can find the Force by the relationship, q*E. Next, we can use the Force found to find the work done: `dW = F * `ds By the relationship `dW +`dKE = 0, we can then find `dKE, which we combine with KE0 to get KEf. **
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RESPONSE --> totally understood
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20:59:37 If we know the charge transferred between two points, the time and the average power necessary to accomplish the transfer, how do we find the potential difference between the points?
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RESPONSE --> Potential difference is work divided charge. power = work divided time.
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20:59:42 ** The potential difference is found from the work done and the charge. Potential difference, or voltage, is work / charge, in Joules / Coulomb. We find the work from the power and the time, since power = work / time. **
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RESPONSE --> ok
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21:02:14 Explain how we can use the flux picture to determine the electric field due to a point charge Q at a distance r from the charge.
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RESPONSE --> electrostatic flux = 4 'pi k Q Since the 'pi is known as well as the constant k we can solve for the flux regardless of the radius.
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21:02:35 STUDENT RESPONSE AND INSTRUCTOR COMMENT: Flux = 4pikQ Flux = area of sphere * electric field = 4 pi r^2 * E k is 9.0 x 10^9 N m^2/C^2 We have 4 pi r^2 * E = 4 pi k Q so E = 4 pi k Q / ( 4 pi r^2) = k Q / r^2 INSTRUCTOR COMMENT: ** Note that the sphere is centered at the charge Q and passes thru the point at distance r so the radius of the sphere is r. Note also that this works because the electric field is radial from Q and hence always perpendicular to the sphere. **
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RESPONSE --> I now see that the radius does make a difference.
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21:03:10 Explain how we can use the flux picture to determine the electric field due to a charge Q uniformly distributed over a straight line of length L, at a distance r << L from that line but not close to either end.
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RESPONSE --> E = flux / area = 4 pi k Q / ( 2 pi r * L) = 2 k Q / L
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21:03:15 ** imagine a circular cylinder around a long segment of the wire; determine the charge on the segment. Total flux is 4 pi k * charge. By the symmetry of the situation the electric field has a very nearly constant magnitude over the curved surface of the cylinder (for an infinite wire the field would be absolutely constant). Almost all of the flux exits the curved surface of the cylinder and is at every point perpendicular to this surface (for an infinite wire all the flux would exit thru the curved surface and would be exactly perpendicular). So you can find flux / area, which is the field. You get E = flux / area = 4 pi k Q / ( 2 pi r * L) = 2 k Q / L. **
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RESPONSE --> ok
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S|~zߩ~ն assignment #023 { Physics II 12-12-2006
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21:10:16 If we know the number of conduction electrons in a wire, the length of the wire and the average drift velocity of the electrons how to we determine the current in the wire?
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RESPONSE --> the velocity gives us 'dL per second then find that the ratio 'dL/L multiply this by 'dL
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21:10:21 GOOD STUDENT SOLUTION: Given: # of electrons, L, vAve of the drift: From the velocity, we find that the electrons will drift a certain `dL per second. We find the ratio of the length `dL/L, and multiply this ratio by the number of electrons in the entire length `dL, to find the number of electrons for that small increment of length and time (1 sec). We can then multiply that number of electrons by the charge of an electron to find the current flowing past a point at any given second. **
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RESPONSE --> ok
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21:11:56 For a given potential difference across two otherwise identical wires, why is the current through the longer wire less than that through the shorter wire?
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RESPONSE --> The potential gradient is higher in the shorter wire.
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21:12:00 ** The potential gradient, which is the potential difference per unit of length, will be higher for the shorter length. The potential gradient is the electric field, which is what exerts the accelerating force on the electrons. So in the shorter wire the electrons are accelerated by a greater average net force and hence build more velocity between collisions. With greater average drift velocity, more electrons therefore pass a given point in a given time interval. **
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RESPONSE --> ok
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21:12:39 For a given potential difference across two otherwise identical wires, why is the current through the thicker wire greater than that through the thinner wire?
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RESPONSE --> There can be more electrons present in the thicker wire than the thinner wire.
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21:12:44 ** The key is that more electrons are available per unit length in the thicker wire. The potential gradient (i.e., the electric field) is the same because the length is the same, so more electrons respond to the same field. GOOD STUDENT SOLUTION: If we know the diameters of both the wires (d1 and d2), we know that the cross-sectional area of the second diameter is (d2/d1)^2 times the cross-sectional area of the first wire. This means that the second wire will have (d2/d1)^2 times as many charge carriers per unit length. This also means that the current in the second wire will be (d2/d1)^2 times that of the first. Therfore the thicker wire will have a greater current.
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RESPONSE --> ok
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21:13:23 If we know the length of a uniform wire and the potential difference between its ends, how do we calculate the average net force exerted on a conduction electron within the wire?
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RESPONSE --> Force is work done divided by the distance that it is done over.
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21:13:31 STUDENT RESPONSE WITH INSTRUCTOR COMMENT: First determine the work done and then divide it by the distance to get the average net force INSTRUCTOR COMMENT: The work done on an electron is the product of its charge and the potential difference. Having this information we can then do as you indicate. GOOD STUDENT SOLUTION: From the charge and voltage (potential difference) we can find how much work is done over the entire length of the wire. We multiply the voltage by the charge to get work done over the full length. We know that F = W/s, so we can divide the work we found by the distance and this will give us the amount of force.
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RESPONSE --> understand
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21:16:34 If we know the voltage and the resistance in a circuit, how do we find the current, and how do we use this result to then reason out the power required to maintain the current?
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RESPONSE --> I = V/R using this equation and solve for I and multiply by the current
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21:16:38 ** Reasoning in terms of units: Power is work per unit of time. P = `dW / `dt, in J / s. Current is voltage / resistance. I = V / R, in C / s. Voltage is work per unit of charge: V = `dW / Q, in J / C. If we multiply voltage in J / C by current in C / s we get power in J / s. CORRECT STUDENT SOLUTION: you find the current by solving the equation I= V/R for I which divided the volts by the resistance of a circuit and to maintain it you multiply the volts by the current determined before.
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RESPONSE --> ok
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21:19:51 Describe the effect of a magnetic field on a current. Note the relative directions of the magnetic field, the current and the force exerted on the current. Note whether a sustained current experiences a sustained force.
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RESPONSE --> B = k' I * 'dL/r^2
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21:19:54 ** A uniform magnetic field B oriented perpendicular to a current I in a straight current carrier of length L exerts a force equal to I * L * B on the current. This force is perpendicular to the magnetic field and to the current by the right-hand rule where I is crossed with B. If B and the current make angle theta then the force is I * L * B * sin(theta). Again the right-hand rule applies. **
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RESPONSE --> ok
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fQHMpݻd|l assignment #024 { Physics II 12-12-2006
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21:21:51 Query problem set 3 #'2 1-6. How do we determine the current in the circuit and the voltage across each resistor when we know the voltage across a series combination of two known resistances?
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RESPONSE --> I = V / R Solve V = I * R
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21:21:57 ** To get the current calculate I = V / R, where R is the sum of the two resistances. To get the voltage across each resistor calculate V = I * R for each resistor. **
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RESPONSE --> ok
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21:22:25 How do we determine the current and voltage across each resistor when we know the voltage across a parallel combination of two known resistances?
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RESPONSE --> I = V / R
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21:22:33 ** The voltage across both resistors is the same and is equal to the voltage across the combination. The current in each resistor is calculated by I = V / R. The total current is the sum of the two currents. **
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RESPONSE --> understood
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21:26:06 A series circuit contains a capacitor of known capacitance and a resistor of known resistance. The capacitor was originally uncharged before the source voltage was applied, and is in the process of being charged by the source. If we know the charge on the capacitor, how do we find the current through circuit?
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RESPONSE --> voltage is equal to the charge voltage across the resistor = difference between source voltage and capacitor voltage divide voltage by resistance to get current
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21:26:14 The voltage across the capacitor is equal to the charge divided by capacitance. The voltage across the capacitor opposes the voltage of the source. Since the voltage drop around the complete circuit must be zero, the voltage across the resistor is the difference between source voltage and the voltage across the capacitor. Dividing the voltage across the resistor by the resistance we obtain the current thru the circuit.
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RESPONSE --> understood
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21:26:47 If we know the capacitance and initial charge on a capacitor in series with a resistor of known resistance then how to we find the approximate time required for the capacitor to discharge 1% of its charge through the circuit?
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RESPONSE --> not sure
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21:26:55 ** From capacitance and initial charge we find the voltage. From the voltage and the resistance we find the current. We take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge. } This result is a slight underestimate of the time required since as the capacitor discharges the current decreases. **
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RESPONSE --> understood
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