Electric Fields - AP Physics 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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Question

Two electrons are deep in space and apart. Determine the magnitude of the electric field at one electron due to the other.

Answer

Using

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

Plugging in values:

$E=9.010^9\frac{({-1.610^{-19}})}{1^2}$

$E=-1.4410^{-9}\frac{N}{C}$

$|E|=1.4410^{-9}\frac{N}{C}$

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Question

Knowing that the electric field away from a point charge is $12: \frac{N}{C}$ , what is the electric field at a distance of away from this charge?

Answer

In this question, we are given the strength of an electric field at a given distance away from a source charge. Then, we're asked to find what the strength of this electric field would be at a different distance away from the source charge.

First, we will need to use the equation for the electric field.

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

Since we're given the strength of the electric field at a distance of away, we can use this information to solve for the value of .

$E_{5}=\frac{kq}{(5: m)^{2}}=12: \frac{N}{C}$

$kq=(12: \frac{N}{C})(5: m)^{2} =300: \frac{N\cdot m^{2}}{C}$

Now that we have the value for , we can use it to find the value of the electric field from the source charge.

$E_{15}=\frac{kq}{(15: m)^{2}}=\frac{300: \frac{N\cdot m^{2}}{C}}{(15: m)^{2}}=1.33: \frac{N}{C}$

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Question

A charge is isolated deep in space. Determine the magnitude and direction of the electric field away.

Answer

Using Coulomb's law:

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

Where

$k=9.010^9\frac{Nm^2}{C^2}$

is the distance, in meters,

and is the charge, in Coulombs.

Plugging in values:

$E=(910^9)\frac{110^{-9}}{10,000^2}$

$E=9*10^{-8}\frac{N}{C}$

The positive sign indicates that the field is pointing away from the charge.

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