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 potion of the illuminated plane is
E perpendicular to xy plane=70
Flux=70*1.5
thanks bob
To calculate the flux through a portion of the illuminated plane, we can use the formula:
Flux = Electric field * Area * cosine(theta)
Where:
- Electric field is the given strength of the electric field (E = 24î + 60j + 70k) N/C
- Area is the area of the portion of the illuminated plane, given as 1.5m²
- cosine(theta) is the angle between the electric field vector and the normal vector of the plane
In this case, since the plane is in the X-Y plane, the normal vector of the plane would be in the Z direction (k). Thus, the angle between the electric field vector and the normal vector would be the angle between the electric field vector and the Z-axis.
To find this angle, we can use the dot product between the electric field vector and the unit vector in the Z direction. The dot product is defined as:
A · B = |A| * |B| * cos(theta)
In this case, A is the electric field vector (E = 24î + 60j + 70k) and B is the unit vector in the Z direction (k = 0î + 0j + 1k).
Taking the dot product:
E · B = (24î + 60j + 70k) · (0î + 0j + 1k)
= 70k
The magnitude of E is given by:
|E| = sqrt((24)^2 + (60)^2 + (70)^2)
Now, we can calculate the flux using the formula:
Flux = Electric field * Area * cos(theta)
We know the electric field is E = 24î + 60j + 70k N/C and the area is given as 1.5m².
cos(theta) is equal to cos(0 degrees) since the angle between the electric field vector and the Z-axis is 0 degrees. The cosine of 0 degrees is 1.
Therefore, the flux through the 1.5m² portion of the illuminated plane is:
Flux = (24î + 60j + 70k) * 1.5 * 1
Please plug in the values to calculate the final answer.