A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
1 answer:
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
Explanation:
From the question we are told that:
Electric field
Distance
At negative plate
Generally the equation for Velocity is mathematically given by
Therefore
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Answer:
<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>
Explanation:
<u>Displacement and Distance</u>
These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:
dtotal=d1+d2+d3+...+dn
This sum is calculated independently of the direction the object moves.
The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.
Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:
d = 25 m + 10 m = 35 m
To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.
The magnitude of the displacement is the distance from (0,0) to (-10,25):
D = 26.93 m
The distance is 35 m and the magnitude of the displacement is 26.93 m