SELECT * FROM question_mgmt as q WHERE id=1138 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=115 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th physics Electric Charges and Field consider a uniform electric field e 3 times 10

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


SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='8' AND ex_no!=0 AND status=1 and id!=1138 ORDER BY views desc, last_viewed_on desc limit 0,10
SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='8' AND ex_no!=0 AND status=1 and id!=1138 ORDER BY last_viewed_on desc limit 0,10

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  • Answered by Ekta Mehta
  • 4 months ago

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