AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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Example Questions

Example Question #81 : Electrostatics

A ball of mass \(\displaystyle 1.9g\)  with \(\displaystyle 1*10^6\)  missing electrons is accelerated with a \(\displaystyle 1.5V\) electric field. Determine the final velocity.

Possible Answers:

\(\displaystyle 1.95*10^{5}\frac{m}{s}\)

\(\displaystyle 2.21*10^{5}\frac{m}{s}\)

None of these

\(\displaystyle 1.95*10^{-5}\frac{m}{s}\)

\(\displaystyle 4.11*10^{5}\frac{m}{s}\)

Correct answer:

\(\displaystyle 1.95*10^{-5}\frac{m}{s}\)

Explanation:

Force in an electric field:

\(\displaystyle F_{Electric Field}=Eq\)

Using conservation of energy, assuming no external work on system:

\(\displaystyle Energy_i=Energy_f\)

\(\displaystyle KE_i+EPE_i=KE_f+EPE_f\)

Definition of electric potential energy:

\(\displaystyle V*q=EPE\)

Combining equations:

\(\displaystyle .5mv^2_i+V_i*q=.5mv^2_f+V_f*q\)

Assuming initial velocity and final electric potential is zero:

\(\displaystyle .5mv^2_f=V_i*q\)

The charge, \(\displaystyle q\) will be equal to the electron charge time the number of electrons missing, \(\displaystyle n*e\)

\(\displaystyle .5mv^2_f=V_i*n*e\)

Converting \(\displaystyle g\) to \(\displaystyle kg\) and plugging in values:

\(\displaystyle .5*.0019*v^2_f=1.5*1*10^{6}*1.6*10^{-19}\)

\(\displaystyle v_f=1.95*10^{-5}\frac{m}{s}\)

Example Question #11 : Electric Potential Energy

If a \(\displaystyle +3\:\mu C\) charged particle moves a distance of \(\displaystyle 7\:mm\) within a \(\displaystyle 2\:V\) electric field, what is the magnitude of change in this particle's electrical potential energy?

Possible Answers:

\(\displaystyle 8.1\cdot10^{-7}\:J\)

\(\displaystyle 4.2\cdot10^{-8}\:J\)

\(\displaystyle 7.9\cdot10^{-6}\:J\)

\(\displaystyle 2.4\cdot10^{-11}\:J\)

\(\displaystyle 3.3\cdot10^{-12}\:J\)

Correct answer:

\(\displaystyle 4.2\cdot10^{-8}\:J\)

Explanation:

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.

\(\displaystyle V=\frac{U}{q}\)

\(\displaystyle U=qV\)

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.

\(\displaystyle V=Ed\)

Combining these expressions, we can obtain our answer.

\(\displaystyle U=qV=qEd\)

\(\displaystyle U=(3\cdot 10^{-6}\:C)(2\:V)(7\cdot 10^{-3}\:m)\)

\(\displaystyle U=4.2\cdot10^{-8}\:J\)

Example Question #81 : Electrostatics

An 8m by 8m square-base pyramid of height 4m is placed in a uniform vertical electric field of strength \(\displaystyle 35 \frac{N}{C}\). What is the total electric flux that goes through the pyramid's four faces? (There is no charge inside the pyramid.)

Possible Answers:

\(\displaystyle 0\frac{N\cdot m^2}{C}\)

\(\displaystyle 360\frac{N\cdot m^2}{C}\)

\(\displaystyle 4480\frac{N\cdot m^2}{C}\)

\(\displaystyle 240\frac{N\cdot m^2}{C}\)

\(\displaystyle 2240\frac{N\cdot m^2}{C}\)

Correct answer:

\(\displaystyle 2240\frac{N\cdot m^2}{C}\)

Explanation:

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 

\(\displaystyle \Phi = \vec{E}\cdot \vec{A}\)

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.

\(\displaystyle \Phi&=E\cdot A\)

\(\displaystyle \phi =35\frac{N}{C}\cdot (8m\cdot 8m)\)

\(\displaystyle \phi = 2240 \frac{N\cdot m^2}{C}\)

Example Question #1 : Gauss's Law

You have a cube with a \(\displaystyle 6\mu C\) 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?

\(\displaystyle \epsilon_0=8.854 \cdot 10^{-12}\)

Possible Answers:

\(\displaystyle 0\frac{N\cdot m^2}{C}\)

\(\displaystyle 1.78\cdot 10^3\frac{N\cdot m^2}{C}\)

\(\displaystyle 6.78\cdot 10^5\frac{N\cdot m^2}{C}\)

\(\displaystyle 2.54\cdot 10^3\frac{N\cdot m^2}{C}\)

\(\displaystyle 1.13\cdot 10^5\frac{N\cdot m^2}{C}\)

Correct answer:

\(\displaystyle 1.13\cdot 10^5\frac{N\cdot m^2}{C}\)

Explanation:

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:

\(\displaystyle \Phi = \vec{E}\cdot\vec{A}=\frac{q_{enclosed}}{\epsilon_0}\)

\(\displaystyle \epsilon_0=8.854\times10^{-12}\)

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.

\(\displaystyle \Phi=\frac{q_{enclosed}}{\epsilon_0}\)

\(\displaystyle \phi = \frac{6\cdot 10^{-6} }{8.854\cdot 10^{-12}}\)

\(\displaystyle \phi = 6.777\cdot 10^{5}\)

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 \(\displaystyle 1.13\cdot 10^{5}\frac{N\cdot m^2}{C}\).

Example Question #1 : Gauss's Law

What is the flux from a \(\displaystyle 12\mu C\) charge inside of a sphere?

\(\displaystyle \epsilon_o=8.854 \cdot 10^{-12}\)

Possible Answers:

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

\(\displaystyle 4.567\cdot 10^5\)

\(\displaystyle 1.355\cdot 10^5\)

\(\displaystyle 1.355\cdot 10^6\)

\(\displaystyle 4.567\cdot 10^6\)

Correct answer:

\(\displaystyle 1.355\cdot 10^6\)

Explanation:

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

\(\displaystyle \Phi=\frac{Q_{enclosed}}{\epsilon_0}\)

The amount of charge enclosed is \(\displaystyle 12\mu C\), so if we divide the charge by epsilon naught, we get the answer, \(\displaystyle 1.355\cdot 10^6\).

Example Question #1 : Gauss's Law

Physics2set1q9

A sphere with a uniform volume charge distribution \(\displaystyle p_{v}= 3\frac{C}{m^3}\) has a radius of 3m. What is the electric field at point C?

Possible Answers:

\(\displaystyle 6.61\pi \cdot 10^{9}\)

\(\displaystyle 20.25\pi \cdot 10^{9}\)

\(\displaystyle 7.2\pi \cdot 10^{9}\)

\(\displaystyle 36\pi \cdot 10^{9}\)

\(\displaystyle 27\pi \cdot 10^{9}\)

Correct answer:

\(\displaystyle 6.61\pi \cdot 10^{9}\)

Explanation:

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.

\(\displaystyle \frac{Q_{tot}}{\epsilon_{0}}= E*A=E(4\pi r^2)\)

We can use this equation to solve for \(\displaystyle E\), but first we need to calculate the total charge.

\(\displaystyle Q_{tot}=4\pi r^2\)

\(\displaystyle Q_{tot}=36\pi\)

Now, plug this into the original equation.

\(\displaystyle \frac{36\pi}{\epsilon_{0}}=E\cdot 4\pi r^2\)

Here, \(\displaystyle r\) is the radius of the Gaussian surface

\(\displaystyle E=\frac{36\pi}{\epsilon_{0}\cdot 4\pi(7)^2}\)

\(\displaystyle E =6.61\pi \cdot 10^9\)

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 \(\displaystyle \sigma =6.5\frac{nC}{m^2}\). What is the strength of the electric field at the surface of this conductor?

Possible Answers:

\(\displaystyle 0.02\frac{N}{C}\)

\(\displaystyle 6.5 \cdot10^{-9}\frac{N}{C}\)

\(\displaystyle 14.8\frac{N}{C}\)

\(\displaystyle 734\frac{N}{C}\)

\(\displaystyle 1.31\cdot 10^{-10}\frac{N}{C}\)

Correct answer:

\(\displaystyle 734\frac{N}{C}\)

Explanation:

Gauss's law tells us that electric field strength is equal to the enclosed charge divided by the vacuum permittivity, \(\displaystyle \epsilon_0\), 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, \(\displaystyle \sigma\), times the area of the sphere.

\(\displaystyle Q=\sigma A\)

\(\displaystyle E=\frac{Q}{\epsilon _0A}\)

\(\displaystyle E=\frac{\sigma A}{\epsilon_0A}\)

\(\displaystyle E=\frac{\sigma }{\epsilon _0}=\frac{6.5\cdot10^{-9}}{8.85\cdot 10^{-12}}=734\frac{N}{C}\)

Example Question #1 : Gauss's Law

Imagine a spherical conducting shell of inner radius \(\displaystyle a\) and outer radius \(\displaystyle b\). If there exists a point charge of 5Q at a radius of less than \(\displaystyle a\) (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:

\(\displaystyle 8Q\)

\(\displaystyle 7Q\)

\(\displaystyle 3Q\)

\(\displaystyle 2Q\)

Correct answer:

\(\displaystyle 8Q\)

Explanation:

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, \(\displaystyle -5Q\) must sit on the inner surface of the conductor(at radius \(\displaystyle a\)). The conductor will put the \(\displaystyle -5Q\) 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 \(\displaystyle -5Q\) 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. \(\displaystyle 8Q+(-5Q)=3Q\)

Example Question #1 : Gauss's Law

A sphere of radius \(\displaystyle 5cm\) contains a charge of \(\displaystyle 66nC\), calculate the electric flux.

Possible Answers:

\(\displaystyle 3.55*10^3V\cdot m\)

\(\displaystyle 6.21*10^3V\cdot m\)

\(\displaystyle 7.45*10^3V\cdot m\)

\(\displaystyle 9.55*10^3V\cdot m\)

\(\displaystyle 2.45*10^3V\cdot m\)

Correct answer:

\(\displaystyle 7.45*10^3V\cdot m\)

Explanation:

Use the equation for electric flux:

\(\displaystyle Flux_E=\frac{Q}{\epsilon_0}\)

Plug in values:

\(\displaystyle Flux_E=\frac{66*10^{-9}}{8.85*10^{-12}}\)

\(\displaystyle Flux_E=7.45*10^3V\cdot m\)

Example Question #151 : Electricity And Magnetism

A \(\displaystyle 15\textup{ nC}\) charge is placed inside of a metal sphere with radius \(\displaystyle 10\textup{ cm}\). Determine the electric field at the surface of the sphere.

Possible Answers:

\(\displaystyle 1.35*10^4\ \frac{\textup{N}}{\textup{C}}\)

\(\displaystyle 8.44*10^4\ \frac{\textup{N}}{\textup{C}}\)

\(\displaystyle 3.95*10^4\ \frac{\textup{N}}{\textup{C}}\)

\(\displaystyle 6.11*10^4\ \frac{\textup{N}}{\textup{C}}\)

None of these

Correct answer:

\(\displaystyle 1.35*10^4\ \frac{\textup{N}}{\textup{C}}\)

Explanation:

Using Electric Field Formula:

\(\displaystyle E=k\frac{q}{r^2}\)

Converting \(\displaystyle nC\) to \(\displaystyle C\), \(\displaystyle cm\) to \(\displaystyle m\) and plugging in values:

\(\displaystyle E=9*10^9 \frac{15*10^{-9}}{.10^2}\)

\(\displaystyle E=1.35*10^4 \frac{N}{C}\)

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