Question

please answer properly ​

Answer 1

Given :

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

Side of the square = 10cm = 0.1m

To find :

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

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

Solution :

(a) The plane of square is parallel to the yz plane. So, angle between the plane and electric field is 0°

Flux through the plane is given by,

$\bf Flux(\phi) = | \vec{E}| A \: cos \theta$

where,

• E denotes Electric field intensity
• A denotes area of square

by substituting all the given values in the equation,

$\dashrightarrow \sf \phi = | \vec{E}| A \: cos \theta$

$\dashrightarrow \sf \phi = 3 \times {10}^{3} \times 0.01 \times cos 0^{ \degree}$

$\dashrightarrow \sf \phi = 30 \: N m^2/C$

Thus, the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane is 30Nm²/C.

(b) Here, the plane makes an angle of 60° with the x-axis so, the angle between them will be 60°

Applying the same formula,

$\dashrightarrow \sf \phi = | \vec{E}| A \: cos \theta$

$\dashrightarrow \sf \phi = 3 \times {10}^{3} \times 0.01 \times cos 60^{ \degree}$

$\dashrightarrow \sf \phi = 3 \times 10 \times \dfrac{1}{2}$

$\dashrightarrow \sf \phi = 30 \times \dfrac{1}{2}$

$\dashrightarrow \sf \phi = 15 \: N m^2/C$

thus, the flux through the same square if the normal to its plane makes a 60° angle with the x- axis is 15Nm²/C.