1. If a surface is a 10 meter by 10 meter plane lying in the y-z plane, what is the vector representing the area of the plane?
2. If a sphere of radius 2 meters is placed in a uniform electric field of 3 N/C, what is the net flux through the sphere?
3. Show how Gauss's Law can be used to derive Coulomb's law for a single charge. Explain any simplifications and draw your Gaussian surface.
100 in the x direction or 100 i
the cross section hit perpendicular to the field is pi r^2
draw sphere centered at origin around charge Q at origin
By symmetry E is constant over surface of sphere at any radius r (surface area 4 pi r^2)
net flux through surface of sphere = E*(4 pi r^2) = Q /eo
so
E = Q/(4 pi r^2 eo)
done
1. To find the vector representing the area of a plane, we can use the cross product of two vectors that lie in the plane. In this case, we have a surface lying in the y-z plane, so we can choose two vectors that are in the y-z plane as well.
Let's choose two vectors: A = (0, 10, 0) and B = (0, 0, 10). These two vectors lie along the sides of the surface in the y and z directions respectively. To find the vector representing the area of the plane, we can take the cross product of A and B:
Area vector = A x B
Taking the cross product of A and B, we get:
Area vector = (10 * 10 - 0 * 0, 0 * 0 - 0 * 10, 0 * 0 - 10 * 10)
= (100, 0, -100)
Therefore, the vector representing the area of the plane is (100, 0, -100).
2. To calculate the net flux through a sphere, we can use Gauss's law. Gauss's law states that the net flux through a closed surface is equal to the enclosed charge divided by the electric constant (ε₀).
In this case, the sphere is in a uniform electric field of magnitude 3 N/C. The electric field is constant and perpendicular to the surface of the sphere at every point, so the electric field lines are parallel to the surface. Therefore, the flux through the sphere will be zero.
Since the flux is zero, the net flux through the sphere is also zero.
3. Gauss's Law can be used to derive Coulomb's law for a single charge by considering a spherical Gaussian surface centered on the charge.
Coulomb's law states that the electric field generated by a point charge is given by the equation:
E = k * (q / r²)
Where E is the electric field, k is the electrostatic constant, q is the charge, and r is the distance from the charge.
To derive this equation using Gauss's Law, we consider a spherical Gaussian surface centered on the charge. The electric field is radially symmetric around the charge, so the electric field will be constant in magnitude on the Gaussian surface.
By applying Gauss's Law, the net flux passing through the Gaussian surface is equal to the enclosed charge divided by ε₀, the electric constant.
Flux = (Enclosed charge) / ε₀
Since the electric field is constant on the surface, we can write the flux as:
Flux = (Electric field magnitude) * (Surface area)
Substituting the expressions for the flux and the electric field magnitude, we get:
(Electric field magnitude) * (4πr²) = (q / ε₀)
Simplifying, we find:
Electric field magnitude = (q / (4πε₀r²))
This is the same equation as Coulomb's law, where k = 1 / (4πε₀).
To draw the Gaussian surface, envision a sphere centered on the charge. The surface of the sphere should enclose the charge completely and be symmetrical around it.