A time-varying potential difference is applied between the plates of a parallel plate capacitor of capacitance . The dielectric constant of the medium between the capacitor plates is 1. It produces an instantaneous displacement current of in the intervening space between the capacitor plates, the magnitude of the rate of change of the potential difference will be ----- .
Official Solution
Correct Option: (1)
The displacement current is related to the rate of change of the charge on the capacitor, which is also related to the rate of change of the potential difference across the plates. The displacement current is given by:
where:
- is the capacitance of the capacitor,
- is the rate of change of the potential difference. Given:
- ,
- . We can solve for :
Solving for :
02
PYQ 2026
medium
physicsID: jee-main
A parallel plate capacitor has a separation between plates of 0.885 mm. It has a capacitance of when the space between the plates is filled with an insulating material of resistivity and resistance . The relative permittivity of the insulating material is . The value of is \underline{\hspace{2cm}}.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
For a medium of given resistivity ( ) and permittivity ( ), the product of its resistance ( ) and capacitance ( ) is a constant that depends only on the material properties, not the geometry. This is derived from the relations and . Step 2: Key Formula or Approach:
The fundamental relationship is:
Step 3: Detailed Explanation:
Given values:
, F, , F/m.
Substituting these into the formula:
Comparing with , we get . Step 4: Final Answer:
The value of is 2.
03
PYQ 2026
medium
physicsID: jee-main
A displacement current of 4.0 A can be set up in the space between two parallel plates of capacitor. The rate of change of potential difference across the plates of the capacitor is nearly . The value of is _________.
1
0.58
2
0.67
3
0.82
4
0.75
Official Solution
Correct Option: (2)
Step 1: Understanding the Question: Displacement current is numerically equal to the conduction current flowing into the capacitor plates. We need to find the rate of change of voltage ( ) given and capacitance . Step 2: Key Formula or Approach:
Step 3: Detailed Explanation: Given and . The relation is:
Rounding to two decimal places, we get . Step 4: Final Answer: The value of is 0.67.
04
PYQ 2026
medium
physicsID: jee-main
A parallel plate air capacitor has a capacitance . When it is half filled as shown in the figure with a dielectric constant , the percentage increase in the capacitance is:
1
2
3
4
Official Solution
Correct Option: (2)
Concept: When a dielectric slab fills part of the separation between plates (layered along thickness), the system behaves like capacitors in series. Original capacitance: Here the dielectric fills half the thickness . Step 1:Capacitance of the air region} Step 2:Capacitance of the dielectric region} Step 3:Equivalent capacitance} Since they are in series: Step 4:Substitute } Step 5:Percentage increase}
05
PYQ 2026
medium
physicsID: jee-main
For the given circuit, find the reading of the ammeter just after the key(s) is closed.
1
1 A
2
3 A
3
A
4
A
Official Solution
Correct Option: (1)
Step 1: Identify the initial condition. When the key is closed, the inductor will act as a short circuit initially (since the inductor's reactance is zero at the moment the circuit is first closed). Thus, the current will initially only be determined by the resistors. Step 2: Analyze the circuit. The circuit consists of three resistors of 9 each in parallel, and the total resistance is given by: Step 3: Apply Ohm's law. Using Ohm's law, the current is: However, as the inductors begin to oppose the change in current, the current will initially be lower. Therefore, just after the key is closed, the current is 1 A. Final Answer:
