A proton accelerates from rest to 3.00x10^6m/s in 1.00x10^-6s in a uniform electric field E. What is the magnitude of the electric field?
acceleration = velocity / time
force = mass * acceleration
field strength = force / charge
To find the magnitude of the electric field, we can make use of the equations of motion for uniformly accelerated motion.
First, we need to find the acceleration of the proton. The equation that relates acceleration, final velocity, initial velocity, and time is:
v = u + at
where:
v = final velocity = 3.00x10^6 m/s
u = initial velocity = 0 m/s (since the proton started from rest)
t = time taken = 1.00x10^-6 s
Rearranging the equation to solve for acceleration (a):
a = (v - u) / t
a = (3.00x10^6 m/s - 0 m/s) / 1.00x10^-6 s
a = 3.00x10^12 m/s^2
Now that we have the acceleration, we can relate it to the electric field using the equation:
a = qE / m
where:
a = acceleration = 3.00x10^12 m/s^2 (which we just calculated)
q = charge of the proton = 1.60x10^-19 C
E = electric field strength (what we're trying to find)
m = mass of the proton = 1.67x10^-27 kg
Rearranging to solve for the electric field (E):
E = (a * m) / q
E = (3.00x10^12 m/s^2 * 1.67x10^-27 kg) / 1.60x10^-19 C
E ≈ 3.12 x 10^4 N/C
So, the magnitude of the electric field is approximately 3.12 x 10^4 N/C.