A radio wave emitter is broadcasting in all directions at 100 MHz (this is in th
ID: 2221295 • Letter: A
Question
A radio wave emitter is broadcasting in all directions at 100 MHz (this is in the FM band). Suppose we choose our definition of time so that the phase constant of this emitter is zero and we define our spatial coordinates so that the emitter is at the origin. Find three locations on the positive x-axis where there are crests at t = 0. A second radio emitter is located on the y-axis at y = 7.35 m. It emits at the same frequency but is half a cycle out of phase with the emitter at the origin. Find the phase of the waves from this second emitter at t = 0 s at each of the locations that you found in part a). Use this to determine whether the interference at each of these points is constructive, destructive, or neither. Now suppose the phase constant of the second emitter is adjusted so that it is in phase with the first emitter. Find the closest point on the x-axis beyond x = 10 m where there will be maximally constructive interference. Note that now that the sources are in phase you don't need to work in terms of phase difference - you can simply work in terms of path length difference. On the negative y-axis, is the interference due to these two sources completely constructive, completely destructive, mostly constructive or mostly destructive? Explain how you know.Explanation / Answer
FOLLOWING IS A SOLVED EXAMPLE JUST PUT DATA IN IT AND GET YOUR ANSWER IT WILLL BE HELPFUL FOR YOU AND PL RATE IF IT IS Draw the x & y axes in standard Cartesian form and plot the points and their charges as given above: q1=5.75 nC at y=+8.00cm q2= 5.75 nC at y=0 q3= 5.75 nC at y=-8cm Mark the point P at x=6.00cm. (1) first, note that all 3 charges are the same in magnitude, and they are all positive; (2) second, note that q1 and q3 are positioned the same distance on either side of the x axis. Because of (1), the 3 electric fields E1, E2, and E3 all point the same direction, which is away from the q-charges [see Source 1] and towards P. Because of (2), the y-components of E1 and E3 at P cancel each other out. From Coulomb's Law, the magnitude of the E field is given by: (3) E = F / q = (k * Qsource * q / r^2) / q = k * Qsource / r^2, where: k = Coulomb's Constant = 8.99e9 N·m^2 / C^2 Qsource = the charge that is the source of the electric field--qi in our case r = the distance between the source charge and the point (P in our case) where you want to measure the value of E. Now, for E1 and E3, the distance r, from Pythagoras, is given by: (4) r^2 = 6^2 + 8^2 = 100cm^2 For E2, which is on the x-axis, r = x = 6cm. Now we can calculate the magnitudes of the 3 E's from (3): (5) |E1| = |E3| = k * Qsource / r^2 = k * Qsource / 100 and (6) |E2| = k * Qsource / r^2 = k * Qsource / 36 The x-component of E1 and E3 at P, again via Pythagoras, is: (7) E1x = E1 * (6 / 10) = 0.6 * E1, and (8) E2x = E2, because there is no y-component for E2 at P. Adding the 3 x-components together: (9) E_x = E1x + E2x + E3x = 2 * E1x + E2x = 2 * (0.6 * E1) + E2 = 1.2 * E1 + E2 = 1.2 * (k * Qsource / 100) + (k * Qsource / 36) = k * Qsource * (1.2 / 100cm^2 + 1 / 36cm^2) = 8.99e9 N·m^2 / C^2 * 5.75 e-9C * (1.2 / 100cm^2 + 1 / 36cm^2) * (1 / 0.0001m^2/cm^2) = 8.99e9 * 5.75 e-9 * (1.2 / 100 + 1 / 36) * (1 / 0.0001) = 2.06 * 10^4 N/C So: E_x = 2.06 * 10^4 N/C E_y = 0 N/C