Electrostatics - AP Physics 2

Card 1 of 444

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Suppose that a charge of $5\textup{ C}$ is moved a distance of $2\textup{ cm}$ from point A to point B while within an electric field. In doing so, $100\textup{ J}$ of work is done. What is the voltage difference between points A and B?

Tap to reveal answer

Answer

For this question, we need to figure out the voltage difference between two points. We're provided with the charge of the particle, the amount of energy put into the process, and the distance traversed by the particle.

First, let's write an equation for voltage.

$\Delta V=\frac{U}{q}$

Where is electrical potential energy, and is the charge of the particle.

This equation describes the change in potential energy that occurs when a given quantity of charge undergoes a displacement while within an electric field. Since we are putting energy into this process to make it occur, and the charge is positive, we know that the voltage change will also be positive; that is, the positively charged particle will move towards the positive terminal of a voltage source and away from the negative terminal.

Plugging in the values given to us, we obtain:

$\Delta V=\frac{100: J}{5: C}=20: \frac{J}{C}$

Notice that we did not need to know the distance that the particle traveled in this case; that information is extraneous.

← Didn't Know|Knew It →

Question

If the area of the plates of a parallel plate capacitor is doubled and the distance between the plates is halved, then the capacitance is .

Tap to reveal answer

Answer

Capacitance is directly proportional to area of the parallel plates and indirectly proportional to the distance between the plates. So $C\sim \frac{A}{d}$ and area is doubled and distance is halved $(\frac{d}{2})$ so the capacitance increases by a factor of 4.

← Didn't Know|Knew It →

Question

What is the electric force between these two point charges?

Tap to reveal answer

Answer

The force between two point charges is shown in the formula below:

$F=k\frac{Q_{1}Q_{2}}{r^2}$ , where $Q_{1}$ and $Q_{2}$ are the magnitudes of the point charges, is the distance between them, and is a constant in this case equal to $8.99\cdot 10^9\frac{N\cdot m^2}{C^2}$

Plugging in the numbers into this equation gives us

$F=8.99\cdot10^9\frac{N\cdot m^2}{C^2}\cdot \frac{(3N\cdot C)(3N\cdot C))}{(3\cdot10^{-4}m)^2}$

$F=9\cdot 10^{-3}N$

← Didn't Know|Knew It →

Question

You have two charges on an axis. One charge of $-2\mu C$ is located at the origin, and the other charge of $-8\mu C$ is located at 4m. At what point along the axis is the electric field zero?

$8.99 \cdot 10^9\frac{N\cdot m^2}{C^2}$

Tap to reveal answer

Answer

The equation for an electric field from a point charge is

$E=\frac{kq}{r^2}$

To find the point where the electric field is 0, we set the equations for both charges equal to each other, because that's where they'll cancel each other out. Let be the point's location. The radius for the first charge would be , and the radius for the second would be .

$\frac{kq_1}{x^2}&=\frac{kq_2}{(4-x)^2}$

$q_1\cdot (4-x)^2= q_2x^2$

$-2\cdot (4-x)^2=-8x^2$

$\frac{4}{3}=x$

Therefore, the only point where the electric field is zero is at $\frac{4}{3}m$ , or 1.34m.

← Didn't Know|Knew It →

Question

A charge of $9\mu C$ is at , and a charge of $4\mu C$ is at . At what point on the x-axis is the electric field 0?

Tap to reveal answer

Answer

To find where the electric field is 0, we take the electric field for each point charge and set them equal to each other, because that's when they'll cancel each other out.

$\frac{kq_1}{r_1^2}=\frac{kq_2}{r_2^2}$

The 's can cancel out.

$\frac{q_1}{r_1^2}&=\frac{q_2}{r_2^2}$

$\frac{9\mu C}{(x)^2}&=\frac{4\mu C}{(4-x)^2}$

$x=\frac{12}{5}$

Therefore, the electric field is 0 at $x = \frac{12}{5}m$ .

← Didn't Know|Knew It →

Question

An electric dipole, with its positive charge above the negative charge, is in a uniform electric field that points to the right, as diagrammed above. What is the net torque and the net force on the dipole in this electric field?

Tap to reveal answer

Answer

Since the net charge of the dipole is zero, the net force will also be zero since . The force on the positive charge on top will be directed to the right since positive charge experiences force in the direction of the electric field. For the negative charge on the bottom, the force will be to the left. Both of the forces contribute to a clockwise torque.

← Didn't Know|Knew It →

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

Tap to reveal answer

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 $-3.36\cdot 10^{-9}\mu C$ :

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

← Didn't Know|Knew It →

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?

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

$N_A=6.022\cdot10^{23}$

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

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.

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

Surface area of sphere:

$4\pi r^2$

Plug in values:

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

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

← Didn't Know|Knew It →

Question

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

Points A and B are point charges.

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

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.

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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:

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

← Didn't Know|Knew It →

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 ?

Tap to reveal answer

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

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

← Didn't Know|Knew It →

Question

Two charges are placed a certain distance apart such that the force that each charge experiences is 20 N. If the distance between the charges is doubled, what is the new force that each charge experiences?

Tap to reveal answer

Answer

To solve this problem, we'll need to utilize the equation for the electric force:

$F_{e}=k\frac{Q_{1}Q_{2}}{r^{2}}$

We're told that the force each charge experiences is 20 N at a certain distance, but then that distance is doubled. Thus, the new electric force will be:

$F_{e}^{}=k\frac{Q_{1}Q_{2}}{(2r)^{2}}=k\frac{Q_{1}Q_{2}}{4r^{2}}=\frac{F_{e}}{4}=\frac{20N}{4}=5N$

← Didn't Know|Knew It →

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

Tap to reveal answer

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:

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

← Didn't Know|Knew It →

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

Tap to reveal answer

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:

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

← Didn't Know|Knew It →

0

Knew It