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.
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* 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: I’m in university physics
.............................................
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): Self-critique Rating: ********************************************* Question: `qquery university physics 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: Distance for q1 to q3 is sqrt(4^2 + 3^2) = 5 cm Distance for q2 to q3 is 3 cm Using Coulomb’s law to find the magnitude of the two forces on q3: 8.988*10^9*[(5*10^-9*6*10^-9)/.05^2] = 0.00011 N 8.988*10^9*[(2*10^-9*6*10^-9)/.03^2] = 0.00012 N For the charge q1 (0,0): Is 4 cm away in the x-direction and 3 cm away in the y-direction. Out of a magnitude of 5 cm away: x-direction of the force: 4/5*0.00011 = 0.000088 N y-direction of the force: 3/5*0.00011 = 0.000066 N For the charge q2 (4,0): It is 0 cm away in the x-direction and 3 cm away in the y-direction. Fx = 0.000088 N Fy = 0.000066-0.00012 = -0.000054 N
.............................................
Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the same answers as you Self-critique Rating: 4 ********************************************* Question: `qQuery univ phy 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: Magnitude of electric field: E = k*abs(q)/r^2 I am going to plug in each point and the corresponding charge to get the magnitude of the electric field: E = k*q/(y+a)^2 + k*q/(y-a)^2 – 2*q/y^2, combining denominators E = [k*q*y^2*(y-a)^2 + k*q*y^2*(y+a)^2 – 2*q*(y+a)^2*(y-a)^2]/[(y+a)^2*(y-a)^2*y^2] E = 2*k*q*[(y^2+a^2)*y^2 – (y+a)^2*(y-a)^2]/[(y+a)^2*(y-a)^2*y^2] Multiplying out and canceling like terms, you get: E = 2*k*q*[3*a^2*y^2 - a^4] / ( y^2*(y+a)^2*(y-a)^2)
.............................................
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 got the final simplified form of the equation; however, I didn’t do the last simplification with the negligible numbers. I understand the final solution though. Self-critique Rating: 3 ********************************************* Question: `qquery univ 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: I am assuming the above number is a mistake and it should read…. 21.102 First, focus on a thin ring on the disk dQ = 2*pi*r*sigma*dr Use this in the expression for the field due equation: dEx =k*[(2*pi*r*sigm*dr)*x/(x^2+r^2)^3/2] Integrate from r1 to r2: Ex = sigma*x/(2*e0)*[1/sqrt(x^2+r1^2) – 1/sqrt(x^2+r2^2)] The direction of the field is along the x-axis
.............................................
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): I got the same general answer, I didn’t do all the analyse that you did at the end; however, I understand what you are saying in it. Self-critique Rating: 3 "