An electron is launched at a ?=54.7° angle and speed of 8.20E+6m/s from the posi

Question

An electron is launched at a ?=54.7° angle and speed of 8.20E+6m/s from the positive plate of the parallel plate capacitor shown. If the electron lands d=3.04cm away, what is the electric field strength inside the capacitor? What is the minimum spacing between the plates?

An electron is launched at a a-54.70 angle and speed of 8.20E+6m/s from the positive plate of the parallel plate capacitor shown. If the electron lands d-3.04cm away, what is the An electron is launched at a a=54.70 angle and speed of 8.20E+6m/s from the positive plate of the parallel plate capacitor shown. If the electron lands d=3.04cm away, what is the electric field strength inside the capacitor? Submit Answer Tries o/10 What is the minimum spacing between the plates? Submit Answer Tries 0/10 Submit Answer Tries 0/10 0

Explanation / Answer

Electron will undergo a projectile motion

Range = (v^2/a) sin (2*54.7) = 0.0304

or a = 0.9432*v^2/0.0304 = 0.9432*(8.2x10^6)^2/(0.0304) or

a =20.8625 x 10^14 m/s^2;

Force on electron = (9.11 x 10^-31) x 20.8625 x 10^14

or Force=19.005 x10^-16 N=eE

So E = 19.005 x10^-16 /(1.6x10^-19)=11.87 x10^3 N/C .............Ans.

So electric field strength inside the capacitor = 11.87 x10^3 N/CIn order that the electron must not collide with the opposite plate It, distance between them must be greater than the height, H of the projectile path.
v^2 = 2aH or

H = v^2/2a = 8.2 x8.2 x10^12/(2x20.8625x10^14)

=1.611x10^-2 m ......Ans.

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