Consider a uniform electric field E = 3 | Class 12 Physics Chapter Electric Charges and Field, Electric Charges and Field NCERT Solutions

Question:

Consider a uniform electric field E = 3 × 103 îN/C.

(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?

(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

Answer:

(a) Electric field intensity, = 3 × 103 î N/C

Magnitude of electric field intensity, = 3 × 103 N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = s2 = 0.01 m2

The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, θ = 0°

Flux (Φ) through the plane is given by the relation,

Φ =

= 3 × 103 × 0.01 × cos0°

= 30 N m2/C

(b) Plane makes an angle of 60° with the x-axis. Hence, θ = 60°

Flux, Φ =

= 3 × 103 × 0.01 × cos60°

= 15 N m2/C


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Comments

  • vashu tyagi
  • 2020-07-22 12:49:29

thanks, now i am out of cofusion that i had with problems similar to this question


  • Md Danish
  • 2019-06-27 21:31:31

Right ans thanks


  • rigio taditya
  • 2018-07-17 18:29:56

Great answer...thank u and it's easy to understand


  • Neshvin
  • 2018-03-17 10:11:56

How it came 60°.since it is a square it should be 90°is it.then the answer is wrong right


  • Anu
  • 2018-03-02 23:30:10

Useful!


  • Deeksha
  • 2016-04-26 17:01:38

Thank u soo much for this wonderful website!


Comment(s) on this Question

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 15: Consider a uniform electric field E = 3 × 103 îN/C. (a) What is the flux of this fiel....