Note: in parts (a) and (c) there should be an arrow over the \"E\" and the \"F\"

Question

Note: in parts (a) and (c) there should be an arrow over the "E" and the "F" but the custom equation function would not work properly. A very thin glass rod 4 meters long is rubbed all over with a silk cloth. It gains a uniformly distributed charge 1.6X10^-6 C. Two small spherical rubber balloons of radius 1.2 cm are rubbed all over with wool. They each gain a uniformly distributed charge of -8X10^-8 C. The balloons are near the midpoint of the glass rod, with their centers 3 cm from the rod. The balloons are 2 cm apart (4.4 cm between centers). (a) Find the net electric field at the location marked by the X, 0.6 cm to the right of the center of the left balloon. Things to think about: Which objects make nonzero contributions to E field at this location? What are the correct distances from sources to the observation location? What is the direction of E field due to the rod? (The drawing is not to scale; read the problem statement carefully. Assume the +x axis is to the right, the +y axis is up, and the +z axis is out.) E = ________ N/C (b) What approximations did you make, if any? Which apply (1-4)? 1. Neglect polarization of balloons. 2. Use approximate formula for electric field of a charged spherical shell. 3. Neglect polarization of rod. 4. Assume distance to observation location is small compared to length of rod. (c) Next a proton is placed at that same location (marked by the x). What is the force acting on the proton? F = ________ N Note: in parts (a) and (c) there should be an arrow over the E and the F but the custom equation function would not work properly.

Explanation / Answer

Both the rod and the balloon contribute to the electric field at X location. making the apprromation that the distance from the point to the rod is lesser than the length of the rod and the charge is uniformly distributed the electric field due to the rod is        E1 = k (2Q /L ) / r here Q = 1.6x10^-6 L = 4m and r = 3cm from the rod.       E1 = (8.99x10^9)[2(1.6x10^-6) /(4) / 3x10^-2 ]            = 2.3973x10^5 N/C     Electric field due to the ballon is     E2 = = k q / R^2    Here q = -8X10^–8 C , R = 2+1.2+0.6 = 3.8cm = 3.8x10^-2m     E2 = (8.99x10^9)(-8X10^–8)/ (3.8x10^-2m )          = - 4.98x10^5 N/C ---------------------------------------------------------------------- b) 1. Neglect polarization of balloons.
     3. Neglect polarization of rod.
     4. Assume distance to observation location is small compared to length of rod. All these assumption are mentioned earlier. ----------------------------------------------------------------------- c) The force on the proton is given as         F = qE           = (1.6x10^-19C) [2.3973x10^5 N/C , - 4.98x10^5 N/C ]           = [ 3.8356x10^-14 N , 7.968x10^-14 N ]

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