A proton (p) and electron (e) are released when they are 10A(10 Angstroms).Find

Question

A proton (p) and electron (e) are released when they are 10A(10 Angstroms).Find the initial accelerations of each particle, if they were free to move.(1 A = 1 times 10^13 m). a(p) m 1.37 times 10^17 m/s^2, a(e) = 2.53 times 10^20 m/s^2; a(p) = 9.4 times 10^18 m/s^2, a(e) = 6.66 times 10^21 m/s^2; a(p) = 4.5 times 10^20 m/s^2, a(e^) = 7.74 times 10^20 m/s^2; or a(p) = 3.5 times 10^18 m/s^2, a(e^) = 6.2 times 10^21 m/s^2. An electron is held stationary between two infinite, parallel, conducting plates that are separated by 1 cm. As shown, the conducting plates are connected to a 100 V power supply. Calculate the magnitude and direction of the electric field generated between the voltage-biased conducting plates. Assume that the positive y-direction is upward. (1 times 10^2 V/m)cap y; -(1 times 10^4 V/m)cap y; - (500 V/m) cap y; or (1 times 10^6 V/m)cap y.

Explanation / Answer

9)

charge of proton and electron is

e = 1.6 * 10^-19 C

distance r = 10 * 10^-10 m = 10^-9 m

the force between them is

F = k * e*e / r^2

F = 9 * 10^9 * (1.6 * 10^-19)^2 / (10^-9)^2

F = 2.3 * 10^-10 N

acceleration a = F / m

for proton

a(p) = 2.3 * 10^-10 / (1.67 * 10^-27)

a(p) = 1.37 * 10^17 m/s^2

for electron

a(e) = 2.3 * 10^-10 / (9.1 * 10^-31)

a(e) = 2.53 * 10^20 m/s^2

10)

electric field E = V / d

E = 100 / 0.01

E = 1 * 10^4 V/m

the direction is towards bottom plate that is -y direction

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