A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then

capacitance will increase.

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energy stored will decrease.

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electric field will increase.

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Voltage will decrease.

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Solution

The correct option is C electric field will increase.
When the capacitor is connected to dc source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases.
Energy stored $(u = \frac{q^{2}}{2 C})$ will increase
Potential $(V = \frac{q}{C})$ will also increase
Electric field $(E = \frac{V}{D})$ will increase

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Similar questions

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.

Q. The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant

K

is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor.
Choose the correct statements.

A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :

Q. A parallel plate capacitor is connected to a battery and a dielectric plate is inserted into the capacitor. Which quantity will increase?

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected?

(a) The electric field between the plates of the capacitor

(b) The energy stored in thecapacitor

Justify your answer by writing the necessary expressions

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