An electron with a horizontal speed of 2300000 m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown. The magnitude of the electric field between the plates is 160 N/C. The plates are 3 cm long. Calculate the angle to the horizontal, 0 (in degrees), that the electron would pass through the plates with. e |+ + + + + + ETZ || 0 L. X

An electron with a horizontal speed of 2300000 m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown. The magnitude of the electric field between the plates is 160 N/C. The plates are 3 cm long. Calculate the angle to the horizontal, 0 (in degrees), that the electron would pass through the plates with. e |+ + + + + + ETZ || 0 L. X

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter25: Gauss’s Law

Section: Chapter Questions

Problem 55PQ: If the magnitude of the surface charge density of the plates in Figure P25.55 is = 99.5 nC/m2, what...

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Figure P24.16 shows three charged particles arranged in the xy plane at the coordinates shown, with qA = qB = 3.30 nC and qC = 4.70 nC. What is the electric field due to these particles at the origin? FIGURE P24.16
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Figure P24.17 shows a dipole (not drawn to scale). If the positive particle has a charge of 35.7 mC and the particles are 2.56 mm apart, what is the (approximate) electric field at point A located 2.00 m above the dipoles midpoint?
An electric dipole is located along the y axis as shown in Figure P24.48. The magnitude of its electric dipole moment is defined as p = 2aq. (a) At a point P, which is far from the dipole (r a), show that the electric potential is V=kepcosr2 (b) Calculate the radial component Er and the perpendicular component E of the associated electric field. Note that E = (1/r)(V/). Do these results seem reasonable for (c) = 90 and 0? (d) For r = 0? (e) For the dipole arrangement shown in Figure P24.48, express V in terms of Cartesian coordinates using r = (x2 + y2)1/2 and cos=y(x2+y2)1/2 (f) Using these results and again taking r a, calculate the field components Ex and Ey. Figure P24.48
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