Electricity and Magnetism - AP Physics 2

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Question

You have 4 resistors and an ammeter arranged as shown in the diagram below.

An ammeter measures current. The ammeter in this setup reads 0A. What is the resistance of ?

Answer

This setup is called a Wheatstone bridge. It's used to find the resistance of a resistor with an unknown value. When the ammeter reads 0, the two sides are at equipotential, so

$\frac{1}{3}=\frac{2}{R}$ .

Therefore, the resistance of R is $6\Omega$ .

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Question

What is the resistance of a $\textup {1.5m}$ long copper wire with a diameter of $\textup {2.2mm}$ ?

$\rho_{copper}=1.72 * 10^{-8}\Omega\textup{m}$

Answer

Write the following formula to find the resistance of the copper wire.

$R=\rho(\frac{L}{A})$

where $\rho$ is the resistivity in $\Omega \textup{m}$ , $\textup L$ is the length of the wire in $\textup m$ , and $\textup A$ is the cross sectional area of the wire in $\textup{m}^\textup {2}$ .

The resistivity of copper is: $1.72 * 10^{-8} \Omega \textup{m}$ .

$\textup L=1.5 \textup{m}$

$\textup A=\pi r^2=\pi(0.0011m)^2 = 1.2110^{-6} \pi m^2$

Substitute the givens to the resistance formula and solve.

$\textup R=(1.72 10^{-8}\Omega \textup{m})(\frac{1.5 \textup{m}}{1.21*10^-6\pi \textup{m}^2})$

$\textup R= 0.0068\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

Answer

In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

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Question

A conductive rod is moving through a region of magnetic field, directed out of the page as diagrammed above. As a result of its motion, the mobile charge in the rod separates, creating an electric potential across the length of the rod. The length of the rod is 0.12m and the magnitude of the magnetic field is 0.022T. If the rod is moving with velocity $v=17\frac{m}{s}$ , what is the magnitude and direction of the potential from one end of the rod to the other?

Answer

For a conductor moving in a magnetic field, cutting straight across the field lines, a potential is generated $\varepsilon=Blv$ where $\varepsilon$ is the potential or EMF, is the magnetic field strength, and is the conductor's velocity relative to the field.

$\varepsilon=0.022T* 0.12m*17\frac{m}{s} =0.045V$

As the rod moves, the positive charge feels an upward-directed force by the right-hand rule, and the negative a downward force resulting in the top being at a higher potential than the bottom of the rod.

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Question

There is a particle with a charge of $11\mu C$ moving $4.7 \cdot 10^7 \frac{m}{s}$ perpendicular through a magnetic field with a strength of . What is the force on the particle?

Answer

The equation for force on a moving charged particle in a magnetic field is

$F=qv\cdot B$ .

Because the charge is moving perpendicularly through the magnetic field, we don't have to worry about the cross product, and the equation becomes simple multiplication.

$F= (11\mu C)(4.7\cdot 10^7\frac{m}{s})(7T)$

Therefore, the force experienced by the particle is 3619N.

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Question

A conductive rod is moving through a region of magnetic field, as diagrammed above. As a result of its motion, mobile charge carriers in the conductor separate, creating an electric potential across the rod. When, if ever, do the charge carriers cease this motion?

Answer

The separated charge creates a potential $\varepsilon=Blv$ . This potential results in an electric field $E=\frac{\varepsilon}{l}=\frac{Blv}{l}=Bv$ When this induced electric field creates a force equal to the magnetic force on the mobile charge carriers, motion stops. Of course, if an electric circuit is created drawing current from the rod, motion will resume to rebuild the field.

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Question

What is the value of the electric field at point C?

Points A and B are point charges.

Answer

First, let's calculate the electric field at C due to point A.

$E_{1}=k\frac{Q_{1}}{R^2}$

$E_{2}=k\frac{Q_{2}}{R^2}$

We can tell that the net electric field will be in the direction.

$E_{net}=E_{2}-E_{1}$

$E_{net}=\frac{k\cdot 10^{-9}}{1^2}(3-\frac{1}{2})$

$E_{net}= 22.5 \frac{N}{C}$ in the direction.

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Question

What is the electric field away from a particle with a charge of ?

Answer

Use the equation to find the magnitude of an electric field at a point.

$E_{field}=\frac{kq}{r^2}$

$E_{field}=9\cdot10^{9}\frac{N\cdot m^{2}}{C^{2}}\cdot\frac{15mC}{(15m)^{2}}$

Solve.

$E_{field}=600000=6\cdot 10^5\frac{N}{C}$

Since it is a positive charge, the electric field lines will be pointing away from the charged particle.

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Question

You are at point (0,5). A charge of is placed at the origin. What charge would you need to place at (0,-3) to cause there to be no net electric field at your location.

Answer

We will need to use the electric field equation, twice. Because we are given coordinates, we will need to use vector notation.

$E_{total}=E_{1}+E_{2}$

$E_{1}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1}$

$E_{2}=k\frac{q_{2}}{r_{2}^2} \widehat{r_{2}}$

Combine the two equations.

$E_{total}=k\frac{q_{1}}{r_{1}^2} \widehat{r_1} + k\frac{q_{2}}{r_{2}^2} \widehat{r_2}$

Plug in known values.

$0 = \frac{50 nC}{5^2}\frac{<0, 5>}{5}+ \frac{q_2}{8^2}\frac{<0, 8>}{8}$

Note that the charge is positive. This is because the electric field lines point towards the negative charge at the origin, and in order to balance this at your location, the electric field lines of the charge at (0,-3) must be pointing away from the charge.

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Question

In the diagram above where along the line connecting the two charges is the electric potential due to the two charges zero?

Answer

Potential is not a vector, so we just add up the two potentials and set them to each other. The equation for electric potential is:

$V=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r}$

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_A}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{Q_B}{(2-d)}$

If the point we are looking for is distance from , it's from . Cancel all the common terms, then cross-multiply:

$\frac{1}{4\pi \epsilon _{0}}\frac{3\times10^{-9}}{d}=\frac{1}{4\pi \epsilon _{0}}\frac{2\times10^{-9}}{(2-d)}$

$\frac{3}{d}=\frac{2}{(2-d)}$

Since we had associated with , it's from that charge toward the weaker charge.

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Question

In the diagram above, where is the electric field due to the two charges zero?

Answer

Electric field is a vector. In between the charges is where 's field points right and 's field points left, so somewhere in between, the two vectors will add to zero. It will be closer to the weaker charge, , but since field depends on the inverse-square of the distance, it will not be linear, and we'll have to do some math.

First set the magnitudes of the two fields equal to each other. The vectors point in opposite directions, so when their magnitudes are equal, the vector sum is zero.

$E_A=E_B \newline \frac{1}{4\pi _0}\frac{Q_A}{d_A^2}=\frac{1}{4\pi _0}\frac{Q_B}{d_B^2}\newline \frac{1}{4\pi _0}\frac{5\times10^{-9}}{d_A^2}=\frac{1}{4\pi _0}\frac{3\times10^{-9}}{(1-d_A)^{2}} \newline \frac{5}{d_A^{2}}=\frac{3}{(1-d_A)^{2}}$

Many of the terms cancel, making it a bit easier. Now cross multiply and solve the quadratic:

$2d_A^{2}-10d_A +5=0$

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Question

If charge has a value of $-910^{-13}C$ , charge has a value of $810^{-13}C$ , and is equal to , what will be the magnitude of the force experienced by charge ?

Answer

Using coulombs law to solve

$F=k\frac{q_1q_2}{r^2}$

Where:

it the first charge, in coulombs.

is the second charge, in coulombs.

is the distance between them, in meters

is the constant of $910^9\frac{Nm^2}{C^2}$

Converting into

$50cm\frac{1m}{100cm}=.50m$

Plugging values into coulombs law

$F=910^{9}\frac{-910^{-13}810^{-13}}{.50^2}$

$F=-2.5910^{-14}N$

Magnitude will be the absolute value

$|F|=2.5910^{-14}N$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to

Answer

Use the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{-2510^{-9}}{.007^2}$

$\overrightarrow{E}=-4.5910^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=4.59*10^6\frac{N}{C}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to

Answer

Using the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{-810^{-9}}{.003^2}$

$\overrightarrow{E}=810^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=8*10^6\frac{N}{C}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to ?

Answer

Using the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{1910^{-9}}{.0045^2}$

$\overrightarrow{E}=8.4410^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=8.44*10^6\frac{N}{C}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to ?

Answer

Use the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{910^{-9}}{.003^2}$

$\overrightarrow{E}=910^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=9*10^6\frac{N}{C}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to

Answer

Use the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{-4810^{-9}}{.025^2}$

$\overrightarrow{E}=-6.9110^5\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=6.91*10^5\frac{N}{C}$

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Question

Charge has a charge of

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the electric field at the center of due to

Answer

Use the electric field equation:

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

Where is $9.010^9 N\frac{m^2}{C^2}$

is the charge, in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{E}=9.010^9\frac{-3810^{-9}}{.015^2}$

$\overrightarrow{E}=-1.5210^6\frac{N}{C}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{E}|=1.52*10^6\frac{N}{C}$

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Question

What is the electric field strength of a stationary 30C charge at a distance of 80cm away?

$k=8.9875*10^{9}: \frac{N\cdot m^{2}}{C^{2}}$

Answer

To solve this question, we need to recall the equation for electric field strength.

$E=k\frac{q_{source}}{r^{2}}$

Notice that the equation above represents an inverse square relationship between the electric field and the distance between the source charge and the point of space that we are interested in.

Plug in the values given in the question stem to calculate the magnitude of the electric field.

$E=\left(8.987510^{9}: \frac{N\cdot m^{2}}{C^{2}}\right)\frac{30 C}{(0.80 m)^{2}}=4.21310^{11} \frac{N}{C}$

Now that we have determined the magnitude of the electric field, we need to identify which direction it is pointing with respect to the source charge. To do this, we'll need to remember that electric fields point away from positive charges and towards negative charges. Therefore, since our source charge is positive, the electric field will be pointing away from the source charge.

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Question

Charge A and B are $15\textup{ cm}$ apart. If charge A has a charge of $2\textup{ nC}$ and a mass of $2\textup{ mg}$ , charge B has a charge of $4\textup{ nC}$ and a mass of $10\textup{ mg}$ , determine the electric field at A due to B.

Answer

Using electric field formula:

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

Converting to , to and plugging in values:

$E=910^9 \frac{410^{-9}}{.15^2}$

$E=1600 \frac{N}{C}$

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