course Phy 202 Dr. Smith:First off I don't have a lot of what the experiments ask for in my kit given to me such as the wood block, the multimeter, and some PVC pipe. I can not do parts of experiments but i read through and try to understand what is being done. I just can not get any data. I am not sure why these were not given to me.
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20:14:34 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 --> A current I flowing across a voltage V results in power P = I V. Therefore current I = V / R will be associated with power P = (V / R) V = V^2 / R
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20:14:46 ** 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, i understand the process
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20:15:37 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 --> You would use I = V / R, then to find the total current just add the two currents
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20:15:45 ** 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 --> Exactly
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20:17:34 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 --> You use the relationship P = I * V which can be combined with either I = V / R or with V = I * R to yield either P = I * (I * R) = I^2 * R or P = (V / R ) * V = V^2 / R
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20:17:42 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 --> OK
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20:24:34 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 --> Greater voltage or less resistance implies greater current, reflecting I = V / R; resistance is the ratio of voltage to current, reflecting R = V / I; a current I through a resistance R requires a greater voltage drop pfor a greater current and for a greater resistance, reflecting V = I * R, using I = V / R or with V = I * R to yield either P = I * (I * R) = I^2 * R or P = (V / R ) * V = V^2 / R
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20:26:19 ** 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 --> OK, so take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge getting a result is a slight underestimate of the time required since as the capacitor discharges the current decreases.
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ws assignment #025 ]ĕ圯N Physics II 11-06-2005
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21:25:35 Query Gen Phy 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE --> First you find the attractice force using colombs law, i found it to be 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 ) * ( 6 * 10^-6 ) / (1)^2 = 324 * 10^-3 N = .32 Newtons. i am stumped here, i am not sure where to go next in this problem
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21:26:31 ** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
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RESPONSE --> OK i was confused when trying to do this: Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. This caused me not to find the whole answer
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21:26:35 query univ 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?
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RESPONSE -->
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21:26:37 ** The 5 nC charge lies at distance sqrt( 4^2 + 3^2) cm from the 6 nC charge, which is 3 cm from the charge at (4 cm ,0). The force exerted on the 6 nC charge by the two respective forces have magnitudes .00011 N and .00012 N respectively. The x and y components of the force exerted by the charge at the origin are in the proportions 4/5 and 3/5 to the total charge, giving respective components .000086 N and .000065 N. The force exerted by the charge at (4 cm, 0) is in the negative y direction. So the y component of the resultant are .000065 N - .00012 N = -.000055 N and its x component is .000086 N. **
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21:26:39 Query univ 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)?
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RESPONSE -->
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21:26:41 ** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 so the expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. **
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21:26:44 query univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?
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RESPONSE -->
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21:26:46 ** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the x component of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x / (x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **
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