Electrostatics - AP Physics 2

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Question

You have a neutral balloon. If you were to add 21,000 electrons to it, what would its net charge be?

$e=1.6\cdot 10^{-19}$ = charge of one electron

Answer

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.

$q=-(#\ of\ electrons)(e)$

$q= -(21000)(1.6\cdot 10^{-19})$

$q= -3.36\cdot 10^{-15}$

When you convert the answer to microcoulombs, the answer is $-3.36\cdot 10^{-9}\mu C$ :

$-3.36\cdot 10^{-15} C\cdot\frac{1.0\cdot10^6 \mu C}{1 C}= -3.36\cdot 10^{-9}\mu C$

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Question

You have of water. One mole of water has a mass of $18 \frac{g}{mol}$ , and a single molecule of water contains 10 electrons. What is the total amount of charge contributed by the electrons in the water?

$e=1.6\cdot 10^{-19}$

$N_A=6.022\cdot10^{23}$

Answer

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:

$1.5 kg_{H_2O} \cdot \frac{1000g}{1 kg}=1500 g_{H_2O}$

Then, we can use the provided molar mass of water to calculate the number of moles of water in 1.5kg of water:

$1500 g_{H_2O}\cdot \frac{1 mol_{H_2O}}{18 g_{H_2O}}=83.33 mol_{H_2O}$

Once we know how many moles of water we have, we can use Avogadro's number ( $6.022\cdot10^{23}\frac{molecules}{mol}$ ) to calculate how many molecules of water are in 83.33mol.

$83.33 mol_{H_2O}\cdot\frac{6.022\cdot10^{23} molecules}{1 mol}=5.0181326\cdot10^{25} molecules_{H_2O}$

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.

$5.0181326\cdot10^{25} molecules_{H_2O}\cdot\frac{10e^-}{1 molecule_{H_2O}}=5.0181326\cdot10^{26} e^-$

Finally, we can multiply by the provided charge of an electron to calculate the charge of those electrons.

$5.0181326\cdot10^{26} electrons\cdot\frac{-1.6\cdot10^{-19}C}{1e^-}=-80290121.6\approx-8.03\cdot10^{-7}C$

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Question

Which of the following best represents the charge of an electron?

$1C=6.24*10^{18}e^-$

Answer

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.

$e=\frac{q}{n}$

One coulomb, , consists of $6.24 * 10^{18}$ electrons,

Substitute these two values into the formula.

$e=\frac{q}{n}=\frac{1 \textup{ C}}{6.24 * 10^{18} \textup{electrons}}$

$e= 1.6 * 10^{-19} \frac{\textup{C}}{\textup{1electron}}$

This number represents the electron's fundamental charge.

$1.6 * 10^{-19} {\textup{C}}$

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Question

Imagine you have a neutral balloon. If you remove 16,000 electrons from it, what is the net charge on the balloon?

$e=1.6\cdot 10^{-19}C$

Answer

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 $1.6\cdot 10^{-19}C$ . Therefore, the total charge that was removed is:

$1.6\cdot 10^{-19}C \cdot 16000$

$=2.56\cdot10^{-15}C$

To answer the question, we must remember that if that much charge was removed from the balloon, the balloon will now be negative.

$=-2.56\cdot10^{-15}C$

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Question

A hollow metal sphere of radius has a charge of distributed evenly on the entirety of the surface. Find the surface charge density.

Answer

Surface area of sphere:

$4\pi r^2$

Plug in values:

$145cm^2=4\pi*r^2$

$\frac{19.9}{145}=.137\frac{nC}{cm^2}$

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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.

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

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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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 B due to A.

Answer

Using electric field formula:

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

Converting to , to and plugging in values:

$E=910^9 \frac{210^{-9}}{.15^2}$

$E=800\frac{N}{C}$

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