A proton, initially traveling in the +x-direction with a speed of 5.30×105m/s , enters a uniform electric field directed vertically upward. After traveling in this field for 4.38×10−7s , the proton’s velocity is directed 45 ∘ above the +x-axis. What is the strength the electic field?
To find the strength of the electric field, we need to use the laws of motion and electromagnetism. Here is how you can calculate it:
1. First, let's understand the situation. A proton is initially traveling in the +x-direction with a speed of 5.30×10^5 m/s. It enters a uniform electric field directed vertically upward and after traveling for a certain time, its velocity is directed 45 degrees above the +x-axis.
2. Now, we can use the equation of motion to calculate the acceleration experienced by the proton in the electric field. The equation is:
F = m * a,
where F is the force acting on the proton, m is the mass of the proton, and a is the acceleration.
3. In this case, the only force acting on the proton is the electric force (F_electric). The equation for electric force is:
F_electric = q * E,
where q is the charge of the proton and E is the strength of the electric field.
4. Since the proton is positively charged, q is equal to the elementary charge (e), which is approximately 1.6 × 10^(-19) C.
5. Now, we can substitute the equation for electric force (F_electric) into the equation for acceleration (a):
q * E = m * a.
6. It is given that the proton's velocity is directed 45 degrees above the +x-axis. We can use trigonometry to determine the value of acceleration (a) in the y-direction. The equation is:
a_y = a * sin(45°).
7. We can use the equation of motion again to find the displacement in the y-direction (Δy) after 4.38 × 10^(-7) s:
Δy = v_y * t + (1/2) * a_y * t^2,
where v_y is the initial velocity in the y-direction, t is the time, and a_y is the acceleration in the y-direction.
8. Since the proton starts with an initial velocity in the +x-direction, we can use the equation of uniform motion to find the displacement in the x-direction (Δx) after 4.38 × 10^(-7) s:
Δx = v_x * t,
where v_x is the initial velocity in the x-direction and t is the time.
9. Now, we know that the displacement in the x-direction (Δx) is equal to the displacement in the y-direction (Δy):
Δx = Δy.
10. We can substitute the values from steps 3, 6, 8, and 9 into the equations and solve for the electric field (E).
11. To obtain the magnitude of the electric field, we can use the following equation:
|E| = sqrt(E_x^2 + E_y^2),
where E_x is the electric field component in the x-direction, and E_y is the electric field component in the y-direction.
Using these steps, you can calculate the strength of the electric field in the given scenario.