What is the acceleration of a proton in an electric field of 10^6 N/C?
To find the acceleration of a proton in an electric field, we need to use the formula for the electric force on a charged particle and Newton's second law of motion.
The formula for the electric force exerted on a charged particle is given by:
F = q * E
where:
- F is the electric force on the charged particle,
- q is the charge of the particle,
- E is the electric field strength.
In this case, we have a proton, which has a charge of +e, where e is the elementary charge (approximately 1.602 x 10^-19 Coulombs).
Given that the electric field strength is 10^6 N/C, we can substitute the values into the formula:
F = (1.602 x 10^-19 C) * (10^6 N/C)
Calculating the result:
F = 1.602 x 10^-19 * 10^6
F = 1.602 x 10^-19 * 10^6 = 1.602 x 10^-13 N
So, the electric force on the proton is 1.602 x 10^-13 N.
Now, we can use Newton's second law of motion to determine the acceleration of the proton. The formula is:
F = m * a
where:
- F is the force,
- m is the mass of the object,
- a is the acceleration.
In this case, we have the electric force (F) and the mass of the proton (m), which is approximately 1.67 x 10^-27 kg.
Substituting the values into the formula:
1.602 x 10^-13 N = (1.67 x 10^-27 kg) * a
We can solve this equation for acceleration (a):
a = (1.602 x 10^-13 N) / (1.67 x 10^-27 kg)
Calculating the result:
a = 1.602 x 10^-13 / 1.67 x 10^-27
a ≈ 9.595 x 10^13 m/s^2
Therefore, the acceleration of a proton in an electric field of 10^6 N/C is approximately 9.595 x 10^13 m/s^2.