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3 years ago

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A parallel-plate capacitor is charged by connecting it to a battery. After that the capacitor is disconnected from the battery a

nd a dielectric of dielectric constant κ is inserted. Choose all correct statements about what happens to the quantities describing that capacitor:
The voltage across the capacitor decreases by a factor of 2.The voltage across the capacitor is doubled. The electric field is doubled.The charge on the plates decreases by a factor of 2.The charge on the plates is doubled.

1 answer:

Eduardwww [97]3 years ago

3 0

Answer:

The voltage across the capacitor decreases by a factor of 2.

Explanation:

As we know that capacitor is initially charges by battery connected across it

so we have

$Q = CV$

now capacitor is disconnected

so charge on the capacitor is conserved

Now a dielectric is inserted between the plates of the capacitor

So we will have

$C' = kC$

now the new voltage across the plates of capacitor is given as

$V = \frac{Q}{C'}$

$V' = \frac{CV}{kC}$

$V' = \frac{V}{k}$

so voltage between the plates of capacitor is decreased by factor of "k"

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