The electric potential inside a 10-m-long linear particle accelerator is given b
ID: 2053815 • Letter: T
Question
The electric potential inside a 10-m-long linear particle accelerator is given by
V = 2450 V (4 V/m2) x2, where x is the distance from the left plate along the accelerator tube, as shown in the figure.
(a) Determine an expression for the electric field along the accelerator tube. (Use the following as necessary: x.)
|E(x)|= ? V/m Answer = 8x
(b) A proton is released (from rest) at x = 4 m. Calculate the acceleration of the proton just after it is released. (Enter the magnitude of the acceleration.)
(c) What is the impact speed of the proton when (and if) it collides with the plate? (If the proton does not collide with the plate, enter 0.)
Explanation / Answer
For part a, we will use the equation:
Ex = -(dV)/(dx) (this applies for any direction, just swap the x's out with y's or z's etc.)
Thus:
Ex = -(dV)/(dx) = -(d/dx)(2450 - 4x2) = -(-8x) = 8x
For part (b), we now need to relate the electric field to acceleration.
We know the equations: F = ma
F = qE
Putting these together, we get: ma = qE
Rearranging: a = qE/m = (charge of proton)*(electric field)/(mass of proton)
Thus:
a = (1.6022×1019 C)*(8x)/(1.6726 * 10-27 kg) = (766327873 s-2)(x)
plugging in x = 4m gives us: a = 3.065 * 109 m/s2
Okay, so now we are going back to electric potential for part (c)
If we take our starting point to be x = 4m, and our final point to be x = 10m, then the change in voltage between those 2 points is V = V_f - V_i = (2450 - 4(102)) - (2450 - 4(42)) = -336 Volts
The change in voltage is related to energy by the equation:
U = qV (U stands for potential energy)
We also know that U = -K
Thus: -K = qV
Because the proton starts from rest:
-0.5mvf2 = qV
vf2 = -2qV/m
vf = (-2qV/m)
Plugging the numbers into this equation gives a result of:
vf = 2.54 * 105 m/s
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