- 1(current)
Q. No.A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:
| 1. | non-zero everywhere with a maximum at the imaginary cylindrical surface connecting the peripheries of the plates. |
| 2. | zero between the plates and non-zero outside. |
| 3. | zero at all places. |
| 4. | constant between the plates and zero outside the plates. |
Subtopic: Â Displacement Current |
Level 3: 35%-60%
NEET - 2025
Please attempt this question first.
Hints
Please attempt this question first.
A parallel plate capacitor is charged by connecting it to a battery through a resistor. If \(i\) is the current in the circuit, then in the gap between the plates:
| 1. | A displacement current of magnitude equal to \(i\) flows in the same direction as \(i.\) |
| 2. | A displacement current of magnitude equal to \(i\) flows in the opposite direction to \(i.\) |
| 3. | A displacement current of magnitude greater than \(i\) flows but it can be in any direction. |
| 4. | There is no current. |
Subtopic: Â Displacement Current |
 62%
Level 2: 60%+
NEET - 2024
Hints
To produce an instantaneous displacement current of \(2~\text{mA}\) in the space between the parallel plates of a capacitor of capacitance \(4~\mu\text{F}\), the rate of change of applied variable potential difference \(\left(\dfrac{dV}{dt}\right) \) must be:
1. \( 800~ \text{V} / \text{s} \)
2. \( 500~ \text{V} / \text{s} \)
3. \( 200~ \text{V} / \text{s} \)
4. \( 400 ~\text{V} / \text{s}\)
1. \( 800~ \text{V} / \text{s} \)
2. \( 500~ \text{V} / \text{s} \)
3. \( 200~ \text{V} / \text{s} \)
4. \( 400 ~\text{V} / \text{s}\)
Subtopic: Â Displacement Current |
 80%
Level 1: 80%+
NEET - 2023
Hints
A capacitor of capacitance \(C\) is connected across an AC source of voltage \(V\), given by;
\(V=V_0 \sin \omega t\)
The displacement current between the plates of the capacitor would then be given by:
\(V=V_0 \sin \omega t\)
The displacement current between the plates of the capacitor would then be given by:
| 1. | \( I_d=\dfrac{V_0}{\omega C} \sin \omega t \) | 2. | \( I_d=V_0 \omega C \sin \omega t \) |
| 3. | \( I_d=V_0 \omega C \cos \omega t \) | 4. | \( I_d=\dfrac{V_0}{\omega C} \cos \omega t\) |
Subtopic: Â Displacement Current |
 60%
Level 2: 60%+
NEET - 2021
Hints
Links
A parallel plate capacitor of capacitance \(20~\mu\text{F}\) is being charged by a voltage source whose potential is changing at the rate of \(3~\text{V/s}.\) The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively:
| 1. | zero, zero | 2. | zero, \(60~\mu\text{A}\) |
| 3. | \(60~\mu\text{A},\) \(60~\mu\text{A}\) | 4. | \(60~\mu\text{A},\) zero |
Subtopic: Â Displacement Current |
 72%
Level 2: 60%+
NEET - 2019
Hints
Links
- 1(current)
Q. No.
© 2026 GoodEd Technologies Pvt. Ltd.