i need help w part b and c
A proton is released from rest in a uniform electric field of magnitude 388 N/C.
(a) Find the electric force on the proton.
magnitude
N
direction
(b) Find the acceleration of the proton.
magnitude m/s2
direction
(c) Find the distance it travels in 2.14 µs.
cm
how to find the porton of a atom
Sure, let's break down parts (b) and (c) step by step.
(b) To find the acceleration of the proton, we can use Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. In this case, the net force is the electric force on the proton, and we can use the formula:
a = F/m
Given that the electric force on the proton is 388 N and the mass of a proton is approximately 1.67 × 10^-27 kg, we can plug these values into the formula to find the acceleration:
a = 388 N / (1.67 × 10^-27 kg)
Calculating this value will give you the magnitude of the acceleration.
To find the direction of the acceleration, we need to consider the direction of the electric field. If the electric field is directed from left to right, then the proton, being positively charged, will experience a force in the opposite direction (right to left). This means that the acceleration will also be in the opposite direction to the field.
(c) To find the distance the proton travels in a given time, we can use the kinematic equation:
d = v * t
Here, "d" represents the distance, "v" represents the velocity, and "t" represents the time.
To find the velocity, we need to know the initial velocity of the proton. Since it is released from rest, the initial velocity is zero. Therefore, the equation simplifies to:
d = 0 * t
Since the initial velocity is zero, the proton will not move in the given time of 2.14 µs. Therefore, the distance it travels is zero.
I hope this answers your question. Let me know if you need any further clarification.