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Vsevolod [243]
2 years ago
11

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

o a battery and charged until its plates carry charges Q and - Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will
Physics
1 answer:
scoundrel [369]2 years ago
6 0

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:

C = \frac{Q}{V}  = e_0\frac{A}{d}

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

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:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

So, if we double the distance between the plates, the potential difference will also be doubled.

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