### All AP Physics 2 Resources

## Example Questions

### Example Question #1 : Gauss's Law

An 8m by 8m square-base pyramid of height 4m is placed in a uniform vertical electric field of strength . What is the total electric flux that goes through the pyramid's four faces? (There is no charge inside the pyramid.)

**Possible Answers:**

**Correct answer:**

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 90^{o} angle.

### Example Question #2 : Gauss's Law

You have a cube with a charge in the center. Each of the cube's sides is 12cm long. What is the flux through one of the faces of the cube?

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #1 : Gauss's Law

What is the flux from a charge inside of a sphere?

**Possible Answers:**

There's not enough information to determine the flux of this situation

**Correct answer:**

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, .

### Example Question #1 : Gauss's Law

A sphere with a uniform volume charge distribution has a radius of 3m. What is the electric field at point C?

**Possible Answers:**

**Correct answer:**

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

### Example Question #1 : Gauss's Law

A spherical conductor has a radius of 0.04m. On the surface an amount of charge is evenly distributed with a surface charge density of . What is the strength of the electric field at the surface of this conductor?

**Possible Answers:**

**Correct 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.

### Example Question #6 : Gauss's Law

Imagine a spherical conducting shell of inner radius and outer radius . If there exists a point charge of 5Q at a radius of less than (it sits within the void inside the conducting shell), and the total charge of the conducting shell is 3Q, what is the magnitude of the charge on the outer surface of the shell?

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Gauss's Law

A sphere of radius contains a charge of , calculate the electric flux.

**Possible Answers:**

**Correct answer:**

Use the equation for electric flux:

Plug in values:

### Example Question #1 : Gauss's Law

A charge is placed inside of a metal sphere with radius . Determine the electric field at the surface of the sphere.

**Possible Answers:**

None of these

**Correct answer:**

Using Electric Field Formula:

Converting to , to and plugging in values:

### Example Question #1 : Gauss's Law

Determine the electric field on the surface of a sphere of radius if are contained within.

**Possible Answers:**

**Correct answer:**

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:

### Example Question #1 : Gauss's Law

Determine the electric flux on the surface of a ball with radius with a helium nucleus inside.

**Possible Answers:**

None of these

**Correct answer:**

is considered the flux.

Plugging in values: