The electrostatic force on a small spher | Class 12 Physics Chapter Electric Charges and Field, Electric Charges and Field NCERT Solutions

Welcome to the NCERT Solutions for Class 12 Physics - Chapter Electric Charges and Field. This page offers a step-by-step solution to the specific question from Exercise 1, Question 2: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 2:

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Answer:

 (a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q= − 0.8 μC = − 0.8 × 10−6 C

Electrostatic force between the spheres is given by the relation,

F space equals space fraction numerator q subscript 1 q subscript 2 over denominator 4 straight pi element of subscript 0 straight r squared end fraction space
A n d comma space fraction numerator 1 over denominator 4 straight pi element of subscript 0 end fraction space equals space 9 space cross times space 10 to the power of 9 space N space m squared space C to the power of minus 2 end exponent

Where, ∈0 = Permittivity of free space

r squared space equals space fraction numerator q subscript 1 q subscript 2 over denominator 4 straight pi element of subscript 0 straight F end fraction
space space space space equals fraction numerator 0.4 space cross times 10 to the power of minus 6 end exponent space cross times 8 space cross times 10 to the power of minus 6 end exponent space cross times 9 space cross times 10 to the power of 9 over denominator 0.2 end fraction
space space space space equals space 144 space cross times space 10 to the power of minus 4 end exponent
r space equals space square root of 144 space cross times 10 to the power of minus 4 end exponent end root
space space space equals.12 space m space

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.


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  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Electric Charges and Field.
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  • Practice answering questions within the time limit to improve your speed and accuracy.
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Comments

  • Gaurav
  • May 03, 2023

u have give answer that why taking 8c instead of 0.8


  • Sahla kanz
  • Jul 20, 2020

Super


  • krish
  • Dec 10, 2019

this is very useful to me thanks for this


  • Ankur mishra
  • Aug 22, 2019

Very nice


  • Lalisa
  • Aug 19, 2019

Very helpful


  • Amit
  • Apr 30, 2019

Helpful but u haven't given the reason that why u put 8 instead of .8


  • Swati joshi
  • Apr 26, 2019

It's really very helpful


  • Minjae
  • Apr 24, 2019

Ha! Thanks for this so simple it was and 0.8 or 8 no problem take 9.0 if u use o.8 or vice versa


  • Anmol
  • Apr 10, 2019

Why taking 8 instead of .8


  • Kaamil
  • Mar 27, 2019

Very good n thank you for help


Add Comment

Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge ....