A proton is placed at rest in an electric field of 515 N/C. Calculate the speed of the proton 36.0 ns after being released
force*time=change in moentum
Eq*time=mass*velcityChange
To calculate the speed of the proton after it has been released in the electric field, we can use the equation of motion for uniformly accelerated linear motion:
v = u + at
Where:
v = final velocity (speed of the proton)
u = initial velocity (which is zero, as the proton is initially at rest)
a = acceleration of the proton
t = time
In this case, the acceleration of the proton is due to the electric field. The force experienced by the proton is given by:
F = qE
Where:
F = force experienced by the proton
q = charge of the proton (1 elementary charge)
E = electric field strength (515 N/C)
Since only the electric field is acting on the proton, the force is equal to the mass of the proton multiplied by its acceleration:
F = ma
Therefore, we can rewrite the equation as:
ma = qE
Rearranging the equation, we find:
a = qE/m
Now we have the acceleration of the proton, which we can substitute into the equation of motion:
v = 0 + at
To calculate the speed of the proton 36.0 ns after being released, we need to convert the time to seconds:
t = 36.0 ns = 36.0 × 10^(-9) s
Now we can substitute the values into the equation:
v = 0 + (qE/m) × t
The charge of a proton, q, is 1.6 × 10^(-19) C, and the mass of a proton, m, is approximately 1.67 × 10^(-27) kg.
Substituting these values, we get:
v = 0 + (1.6 × 10^(-19) C × 515 N/C) / (1.67 × 10^(-27) kg) × (36.0 × 10^(-9) s)
Simplifying the equation:
v = 1.6 × 515 × 36 / 1.67
Calculating the result:
v ≈ 556,287 m/s
Therefore, the speed of the proton 36.0 ns after being released in the electric field is approximately 556,287 m/s.