All AP Physics 2 Resources
Example Questions
Example Question #11 : Electric Potential Energy
A proton is placed into an electric potential of , determine the final velocity after it is released.
None of these
Conservation of energy:
Assuming no external work done
Electric potential energy:
Solving for velocity:
Plugging in values:
Example Question #781 : Ap Physics 2
A car is traveling at . It is carrying a charge against an electric field of . Determine how far the car will travel before stopping.
None of these
Force in an electric field:
Using conservation of energy, assuming no external work on system:
Definition of electric potential energy:
Definition of electric potential:
Combining equations:
Assuming final velocity is zero:
is the distance traveled
Converting to
Plugging in values:
Solving for
Example Question #81 : Electrostatics
A ball of mass with missing electrons is accelerated with a electric field. Determine the final velocity.
None of these
Force in an electric field:
Using conservation of energy, assuming no external work on system:
Definition of electric potential energy:
Combining equations:
Assuming initial velocity and final electric potential is zero:
The charge, will be equal to the electron charge time the number of electrons missing,
Converting to and plugging in values:
Example Question #11 : Electric Potential Energy
If a charged particle moves a distance of within a electric field, what is the magnitude of change in this particle's electrical potential energy?
In this question, we're given the charge of a particle, the distance that it travels, and the electric field within which this movement occurs. We're asked to find the magnitude of the change in electrical potential energy that this particle undergoes.
We can begin this problem by writing an expression for the electric potential energy.
Since we have the particle's charge, but not its electric potential, we need to find a way to obtain this term. To do this, we can make use of the distance the particle travels, as well as the electric field.
Combining these expressions, we can obtain our answer.
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.)
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.
Example Question #1 : 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?
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?
There's not enough information to determine the flux of this situation
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?
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?
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 #1 : 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?
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.
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