A charge q=3*10^-6coulumb is placed at the centre of a cube of edge L.Determine the electric flux through each of the cube faces.
There are six faces.
fluxeachside= q/6*epsilionnaught.
To determine the electric flux through each face of the cube, we need to consider the electric field passing through each face.
First, we need to calculate the electric field at each face due to the charge at the center of the cube. The electric field created by a point charge at a distance r from the charge is given by:
E = k * q / r^2,
where E is the electric field, k is the Coulomb's constant (9 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge.
Since the charge is placed at the center of the cube, the distance from the charge to each face is L/2 (half the length of the edge of the cube).
Now let's calculate the electric field magnitude at each face:
E = k * q / (L/2)^2
= (9 x 10^9 N m^2/C^2) * (3 x 10^-6 C) / (L/2)^2
Simplifying further:
E = (2 x 9 x 10^9 N m^2/C^2) * (3 x 10^-6 C) / L^2
= (54 x 10^3 N m^2/C) / L^2
= 54 x 10^3 / L^2 N m^2/C
Once we have the electric field, we can calculate the electric flux through each face using Gauss's law:
Flux = E * A * cos(theta),
where Flux is the electric flux, E is the electric field, A is the area of the face, and theta is the angle between the electric field and the normal vector of the face (which is 0° for a cube face).
The area of each face is L^2, and since the angle between the electric field and the normal vector is 0°, the cos(theta) term is 1.
Therefore, the electric flux through each face is:
Flux = E * A * cos(theta)
= (54 x 10^3 / L^2 N m^2/C) * L^2 * 1
= 54 x 10^3 N m^2/C
So, the electric flux through each face of the cube is 54 x 10^3 N m^2/C.