Distinguish between pressure and force
1 answer:
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Answer:
Stay the same
Explanation:
First of all, let's find how the capacitance of the capacitor changes.
Initially, it is given by
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:
The potential difference across the capacitor is given by
where
Q is the charge on the plates
C is the capacitance
We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:
Now we can analyze the electric field between the plates of the capacitor, which is given by
we said that:
- The voltage has doubled: V' = 2 V
- The distance between the plates has doubled: d' = 2 d
therefore, the new electric field will be
So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:
F = qE
Then the force on the particle will stay the same.
Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle,
the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and
is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values ,
and
we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.