A uniform electric field of magnitude 4.6 ✕ 104 N/C is perpendicular to a square sheet with sides 3.0 m long. What is the electric flux through the sheet (in N · m2/C)? (Enter the magnitude.)

41.4* 10^4 N.m^2/C

Explanation:

given:

E= 4.6 * 10^4 N/C

electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field

then electric flux = ∫ E*n dA

                            = ∫ 4.6 * 10^4 * 3*3

                            = 41.4* 10^4 N.m^2/C

41.4 x 10⁴ Nm²/C

The electric flux Φ, through a planar area A, is defined as the product of the electric field, the area of the plane and the cosine of the angle θ, between a perpendicular vector to the area and the electric field. i.e

Φ = E A cos θ        (i)

From the question;

E = 4.6 x 10⁴ N/C

A = area of the square = l²          [l = length of the square = 3m]

=> A = (3m)² = 9m²

θ = 0°    [since the electric field is perpendicular to the square sheet, then the area vector is parallel to the field]

Substitute these values into equation (i) as follows;

Φ = 4.6 x 10⁴ x 9 x cos 0°

Φ = 4.6 x 10⁴ x 9 x 1

Φ = 41.4 x 10⁴ Nm²/C

Therefore, the electric flux through the sheet is 41.4 x 10⁴ Nm²/C

0.863333

86%

hope this

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.