All AP Physics 2 Resources
Example Questions
Example Question #1 : Gauss's Law
Determine the electric field on the surface of a sphere of radius if are contained within.
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 #10 : Gauss's Law
Determine the electric flux on the surface of a ball with radius with a helium nucleus inside.
None of these
is considered the flux.
Plugging in values:
Example Question #91 : Electrostatics
Tom is given an uncharged, solid sphere of radius . He then decides to charge all the material within a distance from the center uniformly. He gives that part of the sphere a total charge of . He then charges the rest of the material uniformly, giving that part of the sphere a total charge of . Besides the very center point of the sphere, at what distance from the center will there be no electric field?
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
Example Question #1 : Charge Distribution
You have of water. One mole of water has a mass of , and a single molecule of water contains 10 electrons. What is the total amount of charge contributed by the electrons in the water?
Because we're talking about electrons, the answer must be negative. The way to solve this is to find how many electrons are in 1.5 kg of water. First, we need to convert kilograms of water into grams of water:
Then, we can use the provided molar mass of water to calculate the number of moles of water in 1.5kg of water:
Once we know how many moles of water we have, we can use Avogadro's number () to calculate how many molecules of water are in 83.33mol.
Once we know how many molecules of water we have, we can multiply by 10 to figure out how many electrons those molecules represent, since we are told that each water molecule has 10 electrons.
Finally, we can multiply by the provided charge of an electron to calculate the charge of those electrons.
Example Question #3 : Using Avogadro's Number
You have a neutral balloon. If you were to add 21,000 electrons to it, what would its net charge be?
= charge of one electron
None of the other answers is correct
The elemental charge is the magnitude of charge, in Coulombs, that each electron or proton has. Because electrons have a negative charge, don't forget to add a negative sign into the equation.
When you convert the answer to microcoulombs, the answer is :
Example Question #101 : Electrostatics
Which of the following best represents the charge of an electron?
The equation for the quantity of charge is:
where is the charge quantity, represents the number of electrons, and is the charge of an electron, also known as the elementary charge.
Rewrite the equation.
One coulomb, , consists of electrons,
Substitute these two values into the formula.
This number represents the electron's fundamental charge.
Example Question #1 : Charge Distribution
Imagine you have a neutral balloon. If you remove 16,000 electrons from it, what is the net charge on the balloon?
Because this is a neutral balloon, the net charge is equal to the charge the was removed, but opposite in sign. There were 16,000 electrons removed, each of which has a charge of . Therefore, the total charge that was removed is:
To answer the question, we must remember that if that much charge was removed from the balloon, the balloon will now be negative.
Example Question #1 : Charge Distribution
A hollow metal sphere of radius has a charge of distributed evenly on the entirety of the surface. Find the surface charge density.
Surface area of sphere:
Plug in values:
Example Question #1 : Other Electrostatic Concepts
You place a box with square faces of side length 4m in a uniform electric field of strength . There is no charge in the box. What is the total electric flux going through the box?
A way to think of flux is to count the number of electric field lines exiting a shape and subtract from it the number of field lines entering the shape. The only way for there to be more lines exiting the shape than entering the shape (and the only way to have any flux) is when the shape is enclosing charge.
In our problem, the box has no charge in it. Therefore, it has the same amount of field lines leaving it as it does entering it. This means that there is 0 net flux through the box.
Example Question #2 : Other Electrostatic Concepts
A point charge can be found at the center of a metal box that is isolated, ungrounded, and uncharged. Which of the following statements is true?
The electric field inside the box is 0.
The electric field outside the box is 0.
The electric field everywhere outside the box can be calculated as if it was a point charge.
The electric field inside the box is a uniform strength.
The net charge on the outside surface of the box is .
The net charge on the outside surface of the box is .
Because the box is a conductor, the electrons are able to shift when charge is applied. Because the box is isolated, there is nothing else exerting influence over the box.
The electrons on the shell are able to move from the electric field made by , so they group together on the inner surface of the box to make a net charge of . Because worth of electrons moved from the outer surface, the outer surface necessarily has a net charge of now, because it was neutral before.