A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is introduced between the plates. The result is

P.d between the plates increases, charge on the plate decreases

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P.d between the plates decreases, charge remains same

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P.d increases, charge remain constant

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P.d decreases, charge increases

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B P.d between the plates decreases, charge remains same
As dielectric slab does not affect charges:
$V = \frac{Q}{C} = \frac{Q d}{K ϵ_{o} A}$

Suggest Corrections

Similar questions

Q. A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in :

Q. A parallel plate capacitor is connected across a 2 V battery and charged. The battery is then disconnected and a glass slab is introduced between plates. Which of the following pairs of quantities decreases?

The plates of a parallel plate capacitor are charged to $200 V$ and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant $5$ and thickness $4 m m$ is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:

Statement 1 :

A parallel plate capacitor is charged by a battery of voltage

V

. The battery is then disconnected. If the space between the plates is filled with a dielectric, the energy stored in the capacitor will decrease.

Statement 2 :

The capacitance of a capacitor increases due to the introduction of a dielectric between the plates.

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:

Join BYJU'S Learning Program

A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is introduced between the plates. The result is

The correct option is B P.d between the plates decreases, charge remains sameAs dielectric slab does not affect charges:V=QC=QdKϵoA

The correct option is B P.d between the plates decreases, charge remains same
As dielectric slab does not affect charges:
$V = \frac{Q}{C} = \frac{Q d}{K ϵ_{o} A}$