An electron is projected with an initial velocity v 0 = 8.4 × 10^7 m/s along the

Question

An electron is projected with an initial velocity v0 = 8.4 × 10^7 m/s along the y-axis, which is the centerline between a pair of charged plates, as shown in the figure. The plates are 1.0 m long and are separated by 0.10 m. A uniform electric field of magnitude E in the +x-direction is present between the plates. If the magnitude of the acceleration of the electron is measured to be 8.9x10^15 what is the magnitude of the electric field between the plates? (e = 1.6 × 10-19 C, melectron = 9.11 × 10-31 kg)

Explanation / Answer

Force, F = m a

Also, force, F = q E

Equating both equations, we get,

ma = qE

8.9 x 1015 x 9.11 x 10-31 = 1.6 x 10-19 x E

E = 50674.37 N/C

So, electric field between the plates would be 50674.37 N/C

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