Gauss's Law - AP Physics 2

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Question

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

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.

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

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

$Q_{tot}=4\pi r^2$

$Q_{tot}=36\pi$

Now, plug this into the original equation.

$\frac{36\pi}{\epsilon_{0}}=E\cdot 4\pi r^2$

Here, is the radius of the Gaussian surface

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

$E =6.61\pi \cdot 10^9$

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Question

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

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

$\Phi = \vec{E}\cdot \vec{A}$

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.

$\Phi&=E\cdot A$

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

$\phi = 2240 \frac{N\cdot m^2}{C}$

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Question

You have a cube with a $6\mu C$ 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?

$\epsilon_0=8.854 \cdot 10^{-12}$

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:

$\Phi = \vec{E}\cdot\vec{A}=\frac{q_{enclosed}}{\epsilon_0}$

$\epsilon_0=8.854\times10^{-12}$

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.

$\Phi=\frac{q_{enclosed}}{\epsilon_0}$

$\phi = \frac{6\cdot 10^{-6} }{8.854\cdot 10^{-12}}$

$\phi = 6.777\cdot 10^{5}$

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 $1.13\cdot 10^{5}\frac{N\cdot m^2}{C}$ .

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Question

What is the flux from a $12\mu C$ charge inside of a sphere?

$\epsilon_o=8.854 \cdot 10^{-12}$

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

$\Phi=\frac{Q_{enclosed}}{\epsilon_0}$

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

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Question

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

Answer

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

$Q=\sigma A$

$E=\frac{Q}{\epsilon _0A}$

$E=\frac{\sigma A}{\epsilon_0A}$

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

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Question

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?

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.

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Question

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

Answer

Use the equation for electric flux:

$Flux_E=\frac{Q}{\epsilon_0}$

Plug in values:

$Flux_E=\frac{6610^{-9}}{8.8510^{-12}}$

$Flux_E=7.45*10^3V\cdot m$

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Question

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

Answer

Using Electric Field Formula:

$E=k\frac{q}{r^2}$

Converting to , to and plugging in values:

$E=910^9 \frac{1510^{-9}}{.10^2}$

$E=1.35*10^4 \frac{N}{C}$

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Question

Determine the electric field on the surface of a sphere of radius $0.333\textup{ m}$ if $1969\textup{ nC}$ are contained within.

Answer

Using Gauss's law:

$\Phi =\frac{Q}{\epsilon_0}$

$E_{surface}A=\frac{Q}{\epsilon_0}$

Where $E_{surface}$ is the electric field at the surface of the enclosed shape

is the surface area of the shape

is the charge enclosed

$\epsilon_0$ is $8.8510^{-12}\frac{F}{m}$

Solving for $E_{surface}$

$E_{surface}=\frac{Q}{\epsilon_0 A}$

Surface area of a sphere:

$A=4\pi r^2$

Plugging in values:

$E_{surface}=\frac{196910^{-9}}{8.8510^{-12} 4\pi .333^2}$

$E_{surface}=1.610^5\frac{N}{C}$

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Question

Tom is given an uncharged, solid sphere of radius . He then decides to charge all the material within a distance $\frac{R}{2}$ from the center uniformly. He gives that part of the sphere a total charge of $\frac{-Q}{2}$ . 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?

Answer

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 $\frac{q_{enc}}{\epsilon}= \int E\cdot da$

$q_{enc}=0$

$0=E(4\pi r^2)$

All the charge is uniformly distributed, so the charge density of the inner sphere is $\frac{\frac{-Q}{2}}{\frac{4\pi (\frac{R}{2})^3}{3}}=\frac{-3Q}{\pi R^3}$ and on the outer sphere is $\frac{Q}{\frac{4\pi(R^3-(\frac{R}{2})^3)}{3}}=\frac{6Q}{7\pi R^3}$ . 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 $\frac{Q}{2}$ from the outer sphere in order to cancel the $\frac{-Q}{2}$ from the inner sphere.

Because everything is uniformly distributed:

$q=\rhoV$ . is charge, $\rho$ is charge density, and is volume. We need this to equal $\frac{Q}{2}$

$\frac{Q}{2}=\frac{6Q}{7\pi R}V$

$V=\frac{7\pi R^3}{12}$

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.

$\frac{7\pi R^3}{12}=\frac{4\pi(r^3-(\frac{R}{2})^3))}{3}$

This gives

$r\approx.825R$

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Question

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

Answer

$\oint{EdA}=\frac{Q}{\epsilon_0}$

$\oint{EdA}$ is considered the flux.

Plugging in values:

$\oint{EdA}=\frac{21.610^{-19}}{8.8510^{-12}}$

$\oint{EdA}=3.6110^{-8}\frac{N\cdot m^2}{C}$

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