An electron and a proton are each placed at
rest in an electric field of 363 N/C.
What is the velocity of the electron 76.1 ns
after being released? Consider the direction
parallel to the field to be positive. The elementary charge is 1.60218 × 10
−19
C and the
mass of an electron is 9.10939 × 10
−31
kg .
The acceleration of the electron is
a = e*E/m = 6.38*10^13 m/s^2
After time t, the speed is (1/2) a t^2.
In your case, t = 76.1*10^-9 s
To find the velocity of the electron after being released in the electric field, we can use the formula for the force experienced by a charged particle in an electric field:
F = q * E
Where F is the force, q is the charge of the particle, and E is the electric field strength.
In this case, the charge of an electron is -1.60218 × 10^-19 C and the electric field strength is 363 N/C. Therefore, we can calculate the force on the electron:
F = (-1.60218 × 10^-19 C) * (363 N/C)
F = -5.81905 × 10^-17 N
Next, we can use Newton's second law, which states that force is equal to mass multiplied by acceleration, to find the acceleration of the electron:
F = m * a
Rearranging the formula to solve for acceleration:
a = F / m
Where m is the mass of the electron, which is 9.10939 × 10^-31 kg.
a = (-5.81905 × 10^-17 N) / (9.10939 × 10^-31 kg)
a = -6.38 × 10^13 m/s^2
The minus sign indicates that the acceleration is in the opposite direction of the electric field, which is expected for an electron.
Now, to find the velocity of the electron after 76.1 ns (nanoseconds), we can use the equation for uniformly accelerated motion:
v = u + a * t
Where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.
Plugging in the values:
v = 0 + (-6.38 × 10^13 m/s^2) * (76.1 × 10^-9 s)
v ≈ -4.854 × 10^12 m/s
The negative sign indicates that the electron is moving opposite to the direction of the electric field.