The drawing shows an electron entering the lower left side of a parallel plate c

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 1.11  

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 1.11 times 106 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.

Explanation / Answer

accleration of the electron is obtained fron the Kinematic eqn v^2= 2aS

a = v^2/2S = 1.11*1.11*10^12/(2*0.0015)

a = 4.1 *10^14 m/s^2

now usine ma = Eq

E = 9.11*10^-31 * 4.1*10^14/(1.6*10^-19)

E = 2.33 *10^3 N/C

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