Electrostatics Physics Electrostatic Potential and Capacitance Class 12th Physics

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mi>K</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> <mi>K</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mi>K</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <msub> <mrow> <mi>F</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> </math> </span></p>

3 Views|Posted 4 months ago

Asked by Shiksha User

1 Answer

Sort by:

A

Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

In air $F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (k \in_{0})} \frac{q_{1} q_{2}}{r^{2}}$

$F' = \frac{F_{0}}{K}$

Upvote

Thumbs Down Icon

Similar Questions for you

Four metallic plates, each with a surface area A, are placed at a distance d from each other as shown in the figure. If points A and B is connected with a battery of emf V, then the charge on the plate a will be

Alok Kumar Singh

If q₁+ q₂ = 3q, then the value of the q₁/q₂, for which the force of repulsion between q₁ and q₂ is maximum is

Alok Kumar Singh

For force between q₁ and q₂ is maximum, q₁ = q₂

A capacitor of 20 μF is charged to 500 volt and a capacitor 10μF is charged to 200 volt and connected. Then after connecting both of them, the common potential for each will be :–

Alok Kumar Singh

Three charges Q,(+q) and (+q) are placed at the vertices of an equilateral triangle of side ℓ as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to :-

Alok Kumar Singh

A positive charge is moved from low potential point A to a high potential point B. Then the electric potential energy

Alok Kumar Singh

When positive charge moves from low potential to high potential region, work is done against electric field and potential energy increases.

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

|

1.2K

Exams

|

6.9L

Reviews

|

1.8M

Answers

Learn more about...

Physics Electrostatic Potential and Capacitance 2025

Physics Electrostatic Potential and Capacitance 2025

View Exam Details

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

or

See what others like you are asking & answering