A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
2 answers:
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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.