Question

The electric field in a region of space can be expressed by (285i^ + 1.13×10^3j^ + 796k^) N/C. If a circle of diameter 15.0 cm is placed in the xz-plane of this region, what will be the electric flux through the circle?

Answer 1

The electric flux through the circle can be calculated using the formula:

Φ = E * A * cosθ

where Φ is the electric flux, E is the electric field, A is the area of the circle, and θ is the angle between the electric field vector and the normal to the surface (in this case, the y-axis).

The area of the circle is:

A = π * (d/2)^2
A = π * (15.0 cm / 2)^2
A = π * (7.5 cm)^2
A = π * 56.25 cm^2
A = 176.7145 cm^2 (converted to m^2)
A = 0.01767145 m^2

Now we need to find the angle θ between the electric field vector and the normal to the surface (y-axis). We can do this by finding the magnitude of the electric field vector in the xz-plane and then using the dot product.

E_xz = sqrt(E_x^2 + E_z^2)
E_xz = sqrt((285 N/C)^2 + (796 N/C)^2)
E_xz = sqrt(81,225 N^2/C^2 + 633,616 N^2/C^2)
E_xz = sqrt(714,841 N^2/C^2)
E_xz = 845 N/C

Now we can find the dot product between the electric field vector and the normal to the surface (y-axis).

E_y = 1.13×10^3 N/C
E_dot_n = E_y * 1
E_dot_n = 1.13×10^3 N/C

Now we can find the angle θ using the formula:

cosθ = E_dot_n / E
cosθ = 1.13×10^3 N/C / 845 N/C
cosθ = 1.336

Since cosθ is greater than 1, this means that the electric field vector is not passing through the circle in the xz-plane. Therefore, the electric flux through the circle is zero.