Consider a uniform electric field E=3 × 103 hati N / C. What is the flux of this field through a square of side 10 cm if the normal to its plane makes a 60° angle with the x -axis?

Question

Consider a uniform electric field E=3 × 103 hati N / C. What is the flux of this field through a square of side 10 cm if the normal to its plane makes a 60° angle with the x -axis? (A) 10 N C -1 m 2 (B) 15 N C -1 m 2 (C) 20 N C -1 m 2 (D) 25 N C -1 m 2

Tardigrade Admin · Accepted Answer

Here, vecE=3 × 103 hati N / C i.e., field is along +ve direction of x -axis. Surface area, S=(10 cm )2=102 cm 2 =102 × 10-4 m 2=10-2 m 2 When normal to the plane makes an angle of 60° with x -axis, then θ=60° φE=E S cos θ=3 × 103 × 10-2 cos 60° =30 × (1/2)=15 N C -1 m 2

Q. Consider a uniform electric field $E = 3 \times 1 0^{3} \hat{i} N / C$ . What is the flux of this field through a square of side $10 c m$ if the normal to its plane makes a $6 0^{\circ}$ angle with the $x$ -axis?

$10 N C^{- 1} m^{2}$

$15 N C^{- 1} m^{2}$

$20 N C^{- 1} m^{2}$

$25 N C^{- 1} m^{2}$

Q. Consider a uniform electric field E=3×103i^N/C. What is the flux of this field through a square of side 10cm if the normal to its plane makes a 60∘ angle with the x -axis?

10NC−1m2

15NC−1m2

20NC−1m2

25NC−1m2

Here, E=3×103i^N/C i.e., field is along +ve direction of x -axis. Surface area, S=(10cm)2=102cm2 =102×10−4m2=10−2m2 When normal to the plane makes an angle of 60∘ with x -axis, then θ=60∘ ϕE​=EScosθ=3×103×10−2cos60∘ =30×21​=15NC−1m2

Q. Consider a uniform electric field $E = 3 \times 1 0^{3} \hat{i} N / C$ . What is the flux of this field through a square of side $10 c m$ if the normal to its plane makes a $6 0^{\circ}$ angle with the $x$ -axis?

$10 N C^{- 1} m^{2}$

$15 N C^{- 1} m^{2}$

$20 N C^{- 1} m^{2}$

$25 N C^{- 1} m^{2}$