The figure below shows a proton entering a parallel-plate capacitor with a speed

Question

The figure below shows a proton entering a parallel-plate capacitor with a speed of 2.30×105 m/s. The proton travels a horizontal distance x = 5.30 cm through the essentially uniform electric field. The electric field of the capacitor has deflected the proton downward by a distance of d = 0.760 cm at the point where the proton exits the capacitor. (You can neglect the effects of gravity.)

a)Using kinematics, find the vertical acceleration (including sign) of the proton in this electric field.

b)Find the magnitude of the force on the proton.

c)Find the strength of the electric field within the capacitor.

d)Find the speed of the proton when it exits the capacitor.

Explanation / Answer

Here ,

u = 2.3 *10^5 m/s

x = 5.3 cm = 0.053 m

d = 0.760 cm = 7.6 *10^-3 m

a)
t = x/v

t = 0.053/(2.3 *10^5) = 2.304 *10^-7 s

let the vertical acceleration is a

d = 0.50 *a * t^2

-7.6 * 10^-3 = 0.50 * a * (2.304 *10^-7)^2

solving for a

a = -2.86 *10^11 m/s^2

the vertical acceleration is -2.86 *10^11 m/s^2

b) magnitude of force on proton = m * a

magnitude of force on proton = 2.86 *10^11 * 1.67 *10^-27

magnitude of force on proton = 4.78 *10^-16 N

c)

Now, for the electric field E

1.602 *10^-19 * E = 4.78 *10^-16

E = 2983 N/C

the electric field is 2983 N/C

d)
speed of the proton = sqrt((2.3 *10^5)^2 + (2.86 *10^11 * 2.304 *10^-7)^2)

speed of the proton = 2.39 *10^5 m/s

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