A parallel plate capacitor is charged by connecting it to a battery through a resistor. If <math> <mrow> <mi>I</mi> </mrow> </math> is the current in the circuit, then in the gap between the plates:

According to modified Ampere's law ? B . d I = μ 0 ( I C + I D ) <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->For Loop L 1 I C ≠ 0 and I D = 0 <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->For Loop L 2 I C = 0 and I D ≠ 0 <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->Due to KCLIC=ID <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]->

Ask & Answer Question

Physics Gravitation Physics Class 11th Gravitation

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:

Option 1 -

There is no current

Option 2 -

Displacement current of magnitude equal to $I$ flows in the same direction as

Option 3 -

Displacement current of magnitude equal to $I$ flows in a direction opposite to that of

Option 4 -

Displacement current of magnitude greater than / flows but can be in any direction

5 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

According to modified Ampere's law

$? B . d I = μ_{0} (I_{C} + I_{D})$

For Loop $L_{1} I_{C} \neq 0$ and $I_{D} = 0$

For Loop $L_{2} I_{C} = 0$ and $I_{D} \neq 0$

Due to $KCL I_{C} = I_{D}$

<div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754992677phpCZkUbn.jpeg" alt="" width="463" height="139" loading="lazy"></picture></div></div>According to modified Ampere's law <math> <mrow> <mo>? </mo> <mi>B</mi> <mo>.</mo> <mi>d</mi> <mi>I</mi> <mo>=</mo> <msub> <mrow> <mi>μ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>D</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816518673"> </o:OLEObject></xml><! [endif]->For Loop  <math> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> <mo>≠</mo> <mn>0</mn> </mrow> </math> and <math> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>D</mi> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816518675"> </o:OLEObject></xml><! [endif]->For Loop  <math> <mrow> <msub> <mrow> <mi>L</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>C</mi> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math> and <math> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mi>D</mi> </mrow> </msub> <mo>≠</mo> <mn>0</mn> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1816518677"> </o:OLEObject></xml><! [endif]->Due to  <math> <mrow>KCL<msub><mrow><mi>I</mi></mrow><mrow><mi>C</mi></mrow></msub><mo>=</mo><msub><mrow><mi>I</mi></mrow><mrow><mi>D</mi></mrow></msub></mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1030" DrawAspect="Content" ObjectID="_1816518678"> </o:OLEObject></xml><! [endif]->

0 Upvotes 0 Downvotes

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 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

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 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

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: