Consider a uniform electric field E = 3 × 103 II^ N/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? - Physics

Numerical

Consider a uniform electric field E = 3 × 103 hat"I" N/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?

Solution

(a) Electric field intensity, vec"E" = 3 × 10hat"I" N/C

Magnitude of electric field intensity, |vec"E"| = 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 yz plane. Hence, the 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 = |vec"E"|"A"costheta

= 3 × 103 × 0.01 × cos 0°

= 30 N m2/C

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

Flux, Φ = |vec"E"|"A"costheta

= 3 × 103 × 0.01 × cos 60°

= 30 xx 1/2

= 15 N m2/C

Concept: Electric Flux
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APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.15 | Page 47
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 15 | Page 47

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