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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 and 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.

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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

[tex]Q = CV[/tex]

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

[tex]C' = kC[/tex]

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

[tex]V = \frac{Q}{C'}[/tex]

[tex]V' = \frac{CV}{kC}[/tex]

[tex]V' = \frac{V}{k}[/tex]

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

ayfat23

The voltage across the capacitor when a parallel-plate capacitor is charged by connecting it to a battery will decreases by a factor of 2

What is a capacitor?

A capacitor serves as a device that store electrical energy, consisting of two conductors in close proximity.

We were told that parallel-plate capacitor is charged by connecting it to a battery, then this can be expressed as [tex]Q= CV[/tex]

we were also told that  thecapacitor is disconnected from the battery with insertion of dielectric of dielectric constant κ, which can be expressed as [tex]C'= kC[/tex]

Then we can get the voltage across the capacitor as V= Q/C'

Then [tex]V' = VC/kC[/tex]

Hence, voltage across the capacitor is [tex]V/k[/tex] which means it decreases by a factor of 2.

Learn more about capacitor at;

https://brainly.com/question/14883923

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