Question

The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right side. The initial speed of the electron is 7.00 x10 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

Answer 1

Solution: 0.150 cm. K 2.00cm let it takes time t to reach exit point. Motion in xa direction. Since there is no horizontal force in xa direction hone ax= 0 (2 cm) = 7x10 xt {-ių: 7 X10 mis į 2x10-2 Irloo os Motion in y direction. J = Qyt + tay t² (0.150x1072) = Oxt + ļ ayx (x1000 j2 2.258606 - 0+ ļ ayx (A xrollo ) 2.25x10 6x49 2x 10-16 ay = 5.5125 X10 miga since mass we neglect fret of the ė is very effect of small hence gravitational force on = q E = may

e Electric field E = may E = g. 1x10 3 x 5.5125x10 1.6810-19 E = 5.016375 X 1oto .68LOS E = 3-14 NYC As