Obtain the formula for the electric fiel | Class 12 Physics Chapter Electric Charges and Field, Electric Charges and Field NCERT Solutions

Current NCERT chapter

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 30:

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

. 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 30:

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Answer:

Take a long thin wire XY (as shown in the following figure) of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.

Let E be the electric field at point A due to the wire, XY.

Consider a small length element dx on the wire section with OZ = x

Let q be the charge on this piece.

∴ q = λdx

Electric field due to the piece,

However,

The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component. When the whole wire is considered, the component dEsinθ is cancelled. Only the perpendicular component dEcosθ affects point A.

Hence, effective electric field at point A due to the element dx is dE1 .

∴ d E1 = 1/4πε0 x λdx. cos θ/(l2 + x2)                  ...(1)

In ∆AZO, tanθ = x/l ⇒ x = l.tanθ                          ...(2)

On differentiating equation (2), we obtain

dx/dθ = l x sec 2 θ ⇒ dx = l x sec 2 θdθ              ...(3)

From equation (2), we have

x2 + l2 = l2tan2θ + l2 = l2 (tan2 θ + 1) = l2 sec2 θ          ...(4)

Putting equations (3) and (4) in equation (1), we obtain

The wire is so long that θ tends from − π/2 to π/2.

By integrating equation (5), we obtain the value of field E1 as,

 

Therefore, the electric field due to long wire is λ/2πε0 l .


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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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 30: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density &....

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