An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
1 answer:
Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
So, if we double the distance between the plates, the potential difference will also be doubled.
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According to the statements the number of electrons is 150, then
e = 150
But there is a positive charge of +22e, then the number of protons would be
p = 150+172
If the mass of the electrons is
And the mass of the protong is
We have that the total mass of the system would be
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.