Gauss's Law - AP Physics 2

Let's apply Gauss's law to solve this problem. First, we imagine a Gaussian surface that encompasses the sphere shown. The appropriate Gaussian surface to select is a sphere due to the symmetry of the shape. The Gaussian surface has a radius of 7m.

Gauss's law says that the total charge enclosed in a Gaussian surface is the electric field within the surface times the surface.

We can use this equation to solve for , but first we need to calculate the total charge.

Now, plug this into the original equation.

Here, is the radius of the Gaussian surface

Because there is no charge inside the pyramid, the total flux for the entire shape must be 0. Since the field is vertical, there must be an equal but opposite amount of flux from the base of the pyramid as the faces.

Gauss' Law is

We have both the field strength and the area, so we just multiply them together. We don't have to worry about cross-products because the field is hitting the base at a 90o angle.

There is nonzero electric flux going through the cube because it encloses charges, so there's more electric field lines going out than going in.

Gauss' Law:

Because we know the amount of charge enclosed and we know epsilon naught (the permittivity of free space), the area of the cube and the electric field strength is irrelevant; we can just calculate it with the charge.

That gives us the total electric flux. What we want is the flux from a single face. Since there are 6 faces, we can just divide that number by 6 to get our answer. Once we do that, we get .

Gauss's law tells us that electric field strength is equal to the enclosed charge divided by the vacuum permittivity, , and the area of the Gaussian surface. Taking a Gaussian surface at (or just over) the surface of the sphere gives our Gaussian area the same area as the sphere. The amount of charge is the surface charge density, , times the area of the sphere.

The main property of a conductor is that the electric field inside the material is zero. So if there is 5Q point charge within the void of the conductor, must sit on the inner surface of the conductor(at radius ). The conductor will put the here in order to ensure that the electric field inside the material is zero. This follows from Gauss' law that says the strength in the electric field is directly related to how much charge is contained within the Guassian surface. The only way to enclose zero charge while a point charge of 5Q exists is to cancel it out with on the inner surface of the conductor. Also note that charges only can exist on the surfaces of the conductors, not within. So if the overall charge of the shell is 3Q, then a total charge of 8Q must reside on the outer surface of the conducting shell.

Use the equation for electric flux:

Plug in values:

Using Electric Field Formula:

Converting to , to and plugging in values:

Using Gauss's law:

Where is the electric field at the surface of the enclosed shape

is the surface area of the shape

is the charge enclosed

is

Solving for

Surface area of a sphere:

Plugging in values:

We need the charge enclosed by a gaussian surface (in this case, we'll use a sphere because there's spherical symmetry) to be zero. When this happens, there cannot be any electric field as

All the charge is uniformly distributed, so the charge density of the inner sphere is and on the outer sphere is . We only need to use the outer sphere's surface density since the zero electric field point cannot enclose zero charge in the inner sphere since it's only enclosing negative charge.

The distance where there is no electric field will be when the gaussian surface encloses from the outer sphere in order to cancel the from the inner sphere.

Because everything is uniformly distributed:

. is charge, is charge density, and is volume. We need this to equal

This is the volume of just the outer surface needed. To find the distance, we just use the formula for the volume of a thick spherical shell and algebra.

Let be the distance where there will be zero electric field.

This gives

is considered the flux.

Plugging in values:

Let's apply Gauss's law to solve this problem. First, we imagine a Gaussian surface that encompasses the sphere shown. The appropriate Gaussian surface to select is a sphere due to the symmetry of the shape. The Gaussian surface has a radius of 7m.

Gauss's law says that the total charge enclosed in a Gaussian surface is the electric field within the surface times the surface.

We can use this equation to solve for , but first we need to calculate the total charge.

Now, plug this into the original equation.

Here, is the radius of the Gaussian surface

Because there is no charge inside the pyramid, the total flux for the entire shape must be 0. Since the field is vertical, there must be an equal but opposite amount of flux from the base of the pyramid as the faces.

Gauss' Law is

We have both the field strength and the area, so we just multiply them together. We don't have to worry about cross-products because the field is hitting the base at a 90o angle.

There is nonzero electric flux going through the cube because it encloses charges, so there's more electric field lines going out than going in.

Gauss' Law:

Because we know the amount of charge enclosed and we know epsilon naught (the permittivity of free space), the area of the cube and the electric field strength is irrelevant; we can just calculate it with the charge.

That gives us the total electric flux. What we want is the flux from a single face. Since there are 6 faces, we can just divide that number by 6 to get our answer. Once we do that, we get .

A way to visualize flux is the amount of electric field lines leaving a shape minus the amount entering the shape. If there is no charge inside a shape, the flux is zero because an equal number of lines are entering as leaving. In this problem, we have charge inside of the sphere, so the flux is nonzero. The equation for flux given enclosed charge is

The amount of charge enclosed is , so if we divide the charge by epsilon naught, we get the answer, .

Gauss's law tells us that electric field strength is equal to the enclosed charge divided by the vacuum permittivity, , and the area of the Gaussian surface. Taking a Gaussian surface at (or just over) the surface of the sphere gives our Gaussian area the same area as the sphere. The amount of charge is the surface charge density, , times the area of the sphere.

The main property of a conductor is that the electric field inside the material is zero. So if there is 5Q point charge within the void of the conductor, must sit on the inner surface of the conductor(at radius ). The conductor will put the here in order to ensure that the electric field inside the material is zero. This follows from Gauss' law that says the strength in the electric field is directly related to how much charge is contained within the Guassian surface. The only way to enclose zero charge while a point charge of 5Q exists is to cancel it out with on the inner surface of the conductor. Also note that charges only can exist on the surfaces of the conductors, not within. So if the overall charge of the shell is 3Q, then a total charge of 8Q must reside on the outer surface of the conducting shell.

Use the equation for electric flux:

Plug in values:

Using Electric Field Formula:

Converting to , to and plugging in values:

Using Gauss's law:

Where is the electric field at the surface of the enclosed shape

is the surface area of the shape

is the charge enclosed

is

Solving for

Surface area of a sphere:

Plugging in values: