(a) Two insulated charged copper spheres | Class 12 Physics Chapter Electric Charges and Field, Electric Charges and Field NCERT Solutions

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Academic session 2026-27, Chapter 1 in the current curriculum.

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 12:

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

. 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 12:

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer:

(a) Charge on sphere A, qA = Charge on sphere B,

qB = 6.5 × 10−7 C

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

Where, ∈0 = Free space permittivity

fraction numerator 1 over denominator 4 straight pi element of subscript 0 end fraction=9x109N m2 C2

∴

=1.52X10-2N Therefore, the force between the two spheres is 1.52 × 10−2 N.

(b) As given in the question

After doubling the charge of sphere,

charge on sphere A would be,

qA = 2 × 6.5 × 10−7 C

and charge on sphere B would be,

qB = 1.3 × 10−6 C

The distance between the spheres is halved as given:

∴r space equals fraction numerator 0.5 over denominator 2 end fraction equals 0.25 space m

Now,force of repulsion between the two spheres,

F equals fraction numerator q subscript A q subscript B over denominator 4 straight pi element of subscript 0 straight r squared end fraction

equals fraction numerator 9 x 10 to the power of 9 x 1.3 x 10 to the power of minus 6 end exponent x 1.3 x 10 to the power of minus 6 end exponent over denominator left parenthesis 0.25 right parenthesis squared end fraction

= 16 × 1.52 × 10−2

= 0.243 N

Therefore, the force of repulsion between the two spheres is 0.243 N.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • 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.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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  • Swati
  • Jul 06, 2019

Thanks


  • Megha
  • Jun 05, 2019

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  • Jashwant
  • Mar 31, 2019

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  • Anisha
  • Aug 19, 2018

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  • Melvin
  • Apr 10, 2018

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  • Swati
  • Dec 11, 2017

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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 12: (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm....

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