Please can someone show me how to go about this problem? The X-Y plane is illuminated by an electric field of strenght given as E=(24î + 60j + 70k) N/C; the flux through a 1.5m² portion of the illuminated plane is
To find the flux through a portion of the illuminated plane, we first need to understand what flux is.
Flux is a measure of how much of a vector field passes through a given surface. In this case, we have an electric field given by E = (24î + 60j + 70k) N/C, and we want to find the flux through a portion of the illuminated plane with an area of 1.5m².
To calculate flux, we use the formula:
Flux = ∫∫E∙dA
Where E is the electric field and dA is an infinitesimal area vector.
Given that the electric field is constant in this case, we can simplify the formula to:
Flux = E∙A
Where E is the magnitude of the electric field, and A is the area of the portion of the illuminated plane.
To find the magnitude of the electric field, we can use the Pythagorean theorem:
|E| = √(E_x² + E_y² + E_z²)
In this case, E_x = 24, E_y = 60, and E_z = 70. Plugging these values into the formula, we get:
|E| = √(24² + 60² + 70²) = √(576 + 3600 + 4900) = √10076
Now we have the magnitude of the electric field. Next, we multiply it by the area of the portion of the illuminated plane:
Flux = |E| * A
Given that the area A is 1.5m², we can plug in the values:
Flux = √10076 * 1.5 = 100.38 * 1.5 = 150.57 Nm²/C
Therefore, the flux through the 1.5m² portion of the illuminated plane is 150.57 Nm²/C.