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:
(9.64 +- 0.86) m/s^2
Explanation:
The generic motion equation for constant acceleration is
Where
X0: initial position
v0: initial speed
a: acceleration
t: time
If the object has an initial speed of zero, and the frame of reference is set conveniently so that the object initial position is zero, the equation simplifies to:
And the acceleration can be obtained as:
Where x is the distance fallen and a = g.
So, with the data x = (100.0 +- 0.03) mm and t = (144 +- 3) ms we can calculate
For the uncertainty we have to calculate the relative uncertainties first
For the distance (100 * 0.3)/100 = 0.3%
For the time (100 * 3)/144 = 2.08%
For multiplications or divisions the relative uncertainties are added
0.3% + 2.08% + 2.08% = 4.46%
We convert this into absolute uncertainty:
(9.64e-3 * 4.46)/100 = 0.00043 mm/(ms^2)
Finally, this is multiplied by a constant scalar, so:
2 * 0.00043 mm/(ms^2) = 0.00086 mm/(ms^2)
We convert the units
0.86 m/(s^2)
And the measurement is (9.64 +- 0.86) m/s^2
A better method is putting the ball in a ramp instead of a free fall, that way the fall is longer and the effect of time measuring uncertainty is reduced.