Question

Problem 3 An electron enters the lower left side of a parallel plate capacitor and exits at the upper right, as shown. The initial speed of the electron is 7"106 m/s. The capacitor is 2 cm long, the plates are 0.15 cm apart. Find the magnitude of the electric field between the plates, assuming that it is uniform everywhere. -2.00 cm 0.150 cm

Answer 1

Problem 3

When the electron enters the parallel plate capacitors ,the electric field is from positive plate to negative plate.

The electron moves in the direction opposite to electric field and deviates at an angle $\Theta$ .

0310 = 301" Don Problem 3 taro) (0) POL electronts deviate by an angle o FICOS Here the 2 cm E 00150cm "/= 0-150 -0.075 here tan o = → As the electric field is foom +ve to negative, the electron moves opposite to the direction Of £ 2013 and devieates at an angle o. Here, m Vestano el m=mass of e- V = Velocity of e- e = charge of e- e = acm (length of pla Therefore, E = 9.109 x10-31 (0-1804)x (0-075) 1.602176x 10-19 Х 0.02

here "106= 0.1804 m. *9.109x10-3kgx (0.1804m)?x (0.045) 1- 60214 x 10-9 X O 02 m 0.0 2223x10-31 0.03204 x 10-19 0.6938 x 10-12 10-12v/m The magnitude of Electric field 157916 SIS 3d between capacitor plate votos 293) E = 0.6938 x 10-12 r/m CSV Co