Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Q-1.15
NCERT Solutions for Class 12th Physics
Consider a uniform electric field E = 3 × 10³ î 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?
Share
Ans (a).
Electric field intensity, E = 3 x 103 î N/C
Magnitude of electric field intensity, |E| = 3 x 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 = |E| Acos Θ = 3 x 103 x 0.01 x cos 0° = 30 Nm²/C
Ans (b).
Plane makes an angle of 60° with the x – axis. Hence, Θ = 60°
Flux, Φ = |E|Aco5 6 = 3 x 103 x 0.01 x cos 60°
= 30 x 1/2 = 15 Nm2/C