physical constants 0-3.5 points OSColPhys1 19 P016 WA Notes Ask Your T A proton

Question

physical constants 0-3.5 points OSColPhys1 19 P016 WA Notes Ask Your T A proton is acted on by an uniform electric field of magnitude 333 N/C pointing in the negative x direction. The partide is initially at rest. (a) In what direction will the charge move? Select-- (b) Determine the work done by the electric ficld when the particle has moved through a distance of 3.95 cm from its initial position. (c) Determine the change in clectric potential energy ot the chorged parlicde (d) Determine the speed of the charged paiticle m/s

Explanation / Answer

(a)  The charge of a proton is
Q = 1.602 x 10^19 Coulombs

The electric field is given E = 333 N/C in the negative x - direction.  

The force on the particle is F=EQ
F = 333 x 1.602 x 10^-19 = 5.33 x 10^-17 Newtons, in the NEGATIVE X DIRECTION.
Since the force is in this direction, the acceleration will also be in this direction.

Thus, the proton will begin moving in the NEGATIVE X DIRECTION.

(b) The work done is W = integral F dx. Since F is constant here, W = F x.

So if the proton moves 3.95 cm = 0.0395 m,

the work, W = 5.33 x 10^-17 x 0.0395 = 2.105 x 10^-18 Joules.

(c) This is the same as the decrease in the proton's electric potential energy.

So, the change in electric potential energy of the charged particle = 2.105 x 10^-18 J

(d) The acceleration of the proton can be found from
F = ma;
a = F/m
The mass of a proton is
m = 1.673 x 10^27 kg

a = (5.33 x 10^-17) / (1.673 x 10^-27) = 3.18 x 10^10 m/s^2

Now we can use the equations of motion:
v = at
x = a/2 t^2

We can relate v to x:
v = sqrt(2ax) = sqrt(2x3.18 x 10^10 x 0.0395) = 50121 m/s

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