Question

Two parallel plates are separated by a distance of 12.3 mm are
connected to a 9 volt battery. A proton is emitted from positive
plate with an initial speed of 1.62×10^4 m/s

(A) What is the speed of the proton when it is at a potential of
1.03 volts?

(B) At what distance from the negative plate is the proton when it
is at a potential of 1.03 volts?

(C) What is the magitude of the electric field between the
plates?

(D) What is the electric potential of the proton when it is 4.62 mm
from the negative plate?

(E) How fast is the proton moving when it is 4.62 mm from the
negative plate?

Answer #1

A)

kinetic energy gained = electric potential energy lost

(0.5) m v^{2} = q (V)

(0.5) (1.67 x 10^{-27}) v^{2} = (1.6 x
10^{-19}) (9 - 1.03)

v = 3.9 x 10^{4} m/s

b)

d = distance between the plates = 12.3 mm = 0.0123 m

V = potential difference = 9 Volt

E = electric field between the plates = V/d = 9/0.0123 = 731.7 V/m

r = distance from negative plate

E r = 1.03

731.7 r = 1.03

r = 1.40 mm

c)

d = distance between the plates = 12.3 mm = 0.0123 m

V = potential difference = 9 Volt

E = electric field between the plates = V/d = 9/0.0123 = 731.7 V/m

d)

r = 4.62 mm = 0.00462m

V = E r = 731.7 x 0.00462 = 3.38 Volts

(E) (1.6 x 10^{-19})(9 - 3.38) + (1.67 x
10^{-27})(1.62 x 10^{4})^{2} /2 = (1.67 x
10^{-27}) v^{2} /2

v = 3.65 x 10^{4} m/s

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