#$&* course Phy 232 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.
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Given Solution: `aSTUDENT 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. ** STUDENT COMMENT: Faraday explain that it reached out from the charge, so would that be a concentration? It seems to me that the concentration would be near the center of the charge and the field around it would be more like radiation extending outward weakening with distance. INSTRUCTOR RESPONSE That's a good, and very important, intuitive conception of nature of the electric field around a point charge. However the meaning of the field is the force per unit charge. If you know the magnitude and direction of the field and the charge, you can find the magnitude and direction of the force on that charge. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qExplain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane due to a given point charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to determine the magnitude of an electric field with points, we use the formula, F = k*(q1)*(q2) / r^2. q1 is the charge of the point at the origin. If the value of q1 is negative, the direction moves directly towards the origin and if the value of q1 is positive then the direction is directly away from the origin. To find the direction of the displacement and the angle we use the inverse tan function. Arctan(y / x) = angle. If the angle comes out to be negative, adding 180 degrees to that angle will move it into a different quadrant, making it positive. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. ** STUDENT QUESTION Why is it just Q and not Q2? INSTRUCTOR RESPONSE q1 is a charge that's actually present. Q is a 'test charge' that really isn't there. We calculate the effect q1 has on this point by calculating what the force would be if a charge Q was placed at the point in question. This situation can and will be expanded to a number of actual charges, e.g., q1, q2, ..., qn, at specific points. If we want to find the field at some point, we imagine a 'test charge' Q at that point and figure out the force exerted on it by all the actual charges q1, q2, ..., qn. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Principles of Physics and General Physics problem 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the force between two points on a square with side lengths 1 meter long, we can use Coulomb’s Law. F = (k)*(q1)*(q2) / r^2. Since we know that the distance between one corner to another is 1 meter, and that is where the two points lie, we can calculate the force between the two points. F = [(9.0*10^9 N*m^2/C)*(6*10^-6 C)*(6*10^-6 C)] / (1 m)^2 = .324 N on each side length. To calculate the charges going in a diagonal from one point to the next is calculated the same way. The only thing that changes is the value of r, because the distance between the charges now changes. By using triangles, we see that the hypotenuse, or r, is now sqrt((1m)^2 + (1m)^2) = 1.41 m. Now using Coulomb’s equation again, we get a force of F = [(9.0*10^9 N*m^2/C)*(6*10^-6 C)*(6*10^-6 C)] / (1.41 m)^2 = .16 N. The point in the lower left hand corner experiences three forces on it. The first is the force of .324 N upward, the next .324 N to the right, and the final is a force of .16 N to the upper right at a 45 degree angle. After finding the angle associated with the upper right force and applying that to find the x and y components, we see that the x and y components are the same. To find the magnitude we use the formula sqrt(.76^2 + .76^2) = 1.08. To find the angle, we use arctan(.76 / .76) = 45 degrees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I used 45 degrees as my angle instead of 225 degrees though it resulted in the same final angle. I see how you got that and I see how to go about the rest of the problem making that minor adjustment. ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery university physics 21.66 / 21.72 11th edition 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, comparing the 6 nC charge with the -2 nC charge: The distance between the two is 3 cm, or .03 m in the y direction. To find the force between the two points, we use Coulomb’s Law. F = [(9.0*10^9 N*m^2/C)*(=2*10^-9 C)*(6*10^-9 C)] / (.03 m)^2 = -1.2*10^-4 N. Now the 6 nC charge and the 5 nC charge at the origin: The distance between the two points is 4 cm, 04 .04 m in the x direction. To find the force between the two points we again Coulomb’s Law. F = [(9.0*10^9 N*m^2/C)*(5*10^-9 C)*(6*10^-6 C)] / (.04 m)^2 = 1.69*10^-4 N. Since the new point with 6nC charge is directly to the right of the point at the origin, and directly beneath the other point, its x and y component are the ones that have already been calculated. The magnitude of the forces acting on the 6nC charge point is sqrt((-1.2*10^-4 N)^2 + (1.69*10^-4 N)^2) = .00018 N or 1.8*10^-4 N. The angle will be arctan((1.2*10^-4 N) / (1.69*10-4 N) = -35 degrees which is equivalent to 324 degrees. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The -2 nC charge lies 3 cm above the 6 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (-2 * 10^-9 C) ( 6 * 10^-9 C) / (.03 m)^2 = .00012 N. The force between the two charges is a force of attraction, so the direction of the force on the 6 nC charge is the negative y direction. The vector force is thus -.00012 N * `j. The 5 nC charge lies at distance 4 cm from the 6 nC charge, so it exerts force 9 * 10^9 N m^2 / C^2 * (6 * 10^-9 C) ( 5 * 10^-9 C) / (.05 m)^2 = .000108 N, approx... The charges repel, so this force acts in the direction of the vector 4 `i + 3 `j representing the displacement from the origin to the point (4 cm, 3 cm). The unit vector in this direction is easily seen to be 4/5 `i + 3/5 `j = .8 `i + .6 `j. It follows that the force vector is .000108 N ( 4/5 `i + 3/5 `j) = .000086 N * `i + .000065 N * `j. The resultant force is therefore the sum of these two vectors, which is about .000086 N * `i - .000055 N * `j. This vector has magnitude sqrt( (.000086 N)^2 + (-.000055 N)^2 ) = .00011 N, approx., and angle arcTan(-.000055 N / (.000086 N) ) = -33 degrees, approx., or 360 deg - 33 deg = 327 deg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery univ phy 21.78 / 21.80 11th edition 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure how to go about answering this problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 sothe expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I now see how using simple, known expressions and combining them can lead to a well thought out answer. I understand how the magnitude was used and how the charges were used in the equation. I now understand the process of this problem and how it was approached. ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qquery univ 21.104 / 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma. What is a total electric charge on the annulus? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the total charge on the annulus, we use the equation, Q(charge) = area*sigma. The area in this case can also be written out as (pi*r2^2 - pi*r1^2) to take account for the outer and inner radii. I know that there must be a change in position on the x axis because the annulus changes but I am not quite sure on how to calculate that. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 xcomponent 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Setting up an integral was crucial to the problem and made things easier to understand, I just didn’t think to use one in my calculations. I see how the use of the charge of a thin ring was used to start off this problem and how it led to finding the magnitude of the field around it. After determining the magnitude of the field and setting up the integral from one radius to the other, I now understand how the rest of the problem was completed. ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!