Two charged parallel plates have a potential difference of 500 V between them, a
ID: 1493297 • Letter: T
Question
Two charged parallel plates have a potential difference of 500 V between them, as shown. The electric field between the plates can be assumed to be uniform. A proton is fired with a speed of 2.00x105 m/s from the point between the plates where V = 125 V, toward the positive plate.
The charge on a proton is +1.60x10-19 C, and the mass of the proton is 1.67x10-27 kg.
Be sure to show all your work and explain your reasoning in each step.
(a) The plates are 2.50 cm apart. Calculate the acceleration of the proton.
(b) Calculate the total energy of the proton.
(c) Calculate the speed of the proton when it is halfway between the plates.
(d) Calculate the electric potential at the place where the proton turns around.
Also, find the distance, x, of the proton from the negative plate when it turns around.
Explanation / Answer
Here ,
V = 125 V
speed, u = 2 *10^5 m/s
charge , q = 1.6 *10^-19 C
mass , m = 1.67 *10^-27 Kg
part A)
d = 2.5 cm
let the acceleration is a
using second law of motion
m * a = q * V/d
1.67 *10^-27 * a = 1.6 *10^-19 * 125/(0.025)
solving for a
a = 4.79 *10^11 m/s^2
the acceleration of the proton is 4.79 *10^11 m/s^2
part b)
total energy of proton = q * V + kinetic energy
total energy of proton = 1.602 *10^-19 * 125 + 0.5 * 1.67 *10^-27 * (2 *10^5)^2
total energy of proton = 5.34 *10^-17 J
the total energy of proton is 5.34 *10^-17 J
part c)
when the proton at is halfway
proton at is halfway , let the speed is v
v^2 - u^2 = 2 * a * d
v^2 - (2 * 10^5)^2 = -2 * 4.79 *10^11 * 0.025/2
solving for v
v = 1.67 *10^5 m/s
the speed of proton is 1.67 *10^5 m/s
part d)
let the distance is x
v^2 - u^2 = 2 * a * d
0^2 - (2 * 10^5)^2 = -2 * 4.79 *10^11 * x
solving for x
x = 0.0417 m
hence , the proton will not change direction between the plates
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