The Question is from NEET guide , and it was created by the author Mr. ANIL KUMAR.G

Question

A parallel plate capacitor is being charged by a constant current $I$. The area of the plates is $A$ and the separation is $d$. The displacement current density between the plates is:

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: 2 (Option B)
## CONCEPT FRAMEWORK
 **Primary Concept**: Displacement current and its relation to conduction current in a capacitor.
 **Related Topics**: Maxwell's equations, Ampere-Maxwell's law, capacitor charging.
 **Prerequisites**: Understanding of current density, displacement current, and capacitor behavior.
## SOLUTION PATHWAY
**Given Information**:
	\u203b Constant charging current, $I$
	\u203b Area of capacitor plates, $A$
	\u203b Separation between plates, $d$
**Method & Steps**:
 1. **Understanding Displacement Current:** When a capacitor is charging, the electric field between the plates changes with time. This changing electric field constitutes a displacement current, $I_D$.  The displacement current is equal to the conduction current ($I$) in the wires connected to the capacitor.
  Reasoning:  Charge conservation requires the current into the capacitor to be equal to the rate of change of charge on the plates. This rate of change of charge is linked to the changing electric field, which defines the displacement current.
  Formula/Rule: $I_D = I$
 2. **Calculating Displacement Current Density:** Current density ($J$) is the current per unit area. In this case, the displacement current density ($J_D$) is the displacement current divided by the area of the capacitor plates.
  Working: $J_D = \frac{I_D}{A} = \frac{I}{A}$
  Key Point: The displacement current is uniformly distributed across the area between the plates.
 3. **Final Result:** The displacement current density between the plates is $\frac{I}{A}$.
  Result: $\frac{I}{A}$
  Verification: The units are consistent (Amperes per square meter, A/m²), which is the correct unit for current density.
 ## OPTION ANALYSIS
[A] :x: $\frac{I}{2A}$: This is incorrect. It assumes that only half of the current contributes to the displacement current, which is not true.
[B] ✓: $\frac{I}{A}$: This is the correct answer, as explained above.
[C] :x: $\frac{2I}{A}$: This is incorrect. It erroneously doubles the current density.
[D] :x: Zero: This is incorrect. A changing electric field between the capacitor plates always generates a displacement current.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Confusing conduction current with displacement current: Remember that in a capacitor, both currents are equal during charging.
  2.Incorrectly using the area: Use the area of the capacitor plates, not some other area.
 **Quick Tips**:
  1.Remember the relationship $I_D = I$ for a capacitor being charged.
  2.Current density is always current divided by area.

Mr. ANIL KUMAR.G · Answer

$\frac{I}{2A}$

Mr. ANIL KUMAR.G · Answer

The correct Option is "$\frac{I}{A}$"

Mr. ANIL KUMAR.G · Answer

$\frac{2I}{A}$

Mr. ANIL KUMAR.G · Answer

Zero

NEET Questions

Select Subject

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company