Calculate the electric flux linked with a square of side 0.5m held parallel to x-y plane when the electric field 2i+3j*10^3 nc^-1 exist in the region

To find the electric flux linked with a square of side 0.5 m held parallel to the x-y plane in the presence of an electric field, we can use Gauss's law. The electric flux (Φ) through a surface is given by the formula:

Electric Flux Formula

Electric flux is calculated using the equation:

Φ = E · A · cos(θ)

Where:

• Φ is the electric flux.
• E is the magnitude of the electric field.
• A is the area of the surface through which the field lines pass.
• θ is the angle between the electric field and the normal to the surface.

Step-by-Step Calculation

In this case, the electric field is given as 2i + 3j × 10^3 N/C. Since the square is parallel to the x-y plane, we can determine the following:

• The area (A) of the square is calculated as:

A = side × side = 0.5 m × 0.5 m = 0.25 m²

• The electric field vector can be expressed as:

E = 2 N/C in the x-direction and 3 × 10^3 N/C in the y-direction.

Finding the Angle

Since the square is parallel to the x-y plane, the normal vector to the surface is in the z-direction (0, 0, 1). The electric field has components in the x and y directions, which means the angle θ between the electric field vector and the normal to the surface is 90 degrees. Thus, cos(θ) = cos(90°) = 0.

Calculating the Electric Flux

Now, substituting the values into the electric flux formula:

Φ = E · A · cos(θ) = (2i + 3 × 10^3 j) · (0.25 m²) · 0

Since cos(90°) = 0, the entire expression becomes:

Φ = 0

Final Result

The electric flux linked with the square is 0 Nm²/C. This indicates that there is no net electric field passing through the square surface, which is consistent with the fact that the electric field is perpendicular to the area vector of the square.