To increase the charge on the plate of a capacitor implies to:-

Decrease the potential difference between the plates.

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

Decreased the capacitance of the capacitor.

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

Increase the capacitance of the capacitor.

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

Increase the potential difference between the plates.

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

Open in App

Solution

The correct option is D Increase the potential difference between the plates.
Charge stored on the capacitor $Q = C V$
So, $Q \propto V$
Thus to increase the charge stored on the plate of the capacitor, we need to increase the potential difference between the plates.

Suggest Corrections

Similar questions

A parallel-plate capacitor has a capacitance of

4 μ F

. Initial charge on the capacitor is

20 μ C

. Which of the following changes would increase potential difference between its plates to

10 V

Q. A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.
(a) The battery will supply more charge.
(b) The capacitance will increase.
(c) The potential difference between the plates will increase.
(d) Equal and opposite charges will appear on the two faces of the metal plate.

Q. A parallel plate air capacitor is charged to a potential difference of V volt. After disconnecting the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result, the potential difference between the plates

A parallel plate capacitor of plate separation d, plate area A and of capacitance C is charged to a potential difference V by connecting it to a battery. When a dielectric slab of thickness t and k = 2 is introduced between the plates of the capacitor, its capacitance increases to $\frac{3}{2} C$ , the ratio $\frac{t}{d}$ is

Q. The plates of a parallel plate capacitor are given charges

+ 4 Q

and

- 2 Q

. The capacitor is then connected across an uncharged capacitor of same capacitance as first one

(= C)

. Find the final potential difference between the plates of the first capacitor.

Join BYJU'S Learning Program

To increase the charge on the plate of a capacitor implies to:-

The correct option is D Increase the potential difference between the plates.Charge stored on the capacitor Q=CVSo, Q∝VThus to increase the charge stored on the plate of the capacitor, we need to increase the potential difference between the plates.

The correct option is D Increase the potential difference between the plates.
Charge stored on the capacitor $Q = C V$
So, $Q \propto V$
Thus to increase the charge stored on the plate of the capacitor, we need to increase the potential difference between the plates.