AP Physics 2 - Electric Fields

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.

$1600\ \frac{\textup{N}}{\textup{C}}$

None of these

$537\ \frac{\textup{N}}{\textup{C}}$

$611\ \frac{\textup{N}}{\textup{C}}$

$925\ \frac{\textup{N}}{\textup{C}}$

Explanation

Using electric field formula:

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

Converting to , to and plugging in values:

$E=9*10^9 \frac{4*10^{-9}}{.15^2}$

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

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

$6\cdot 10^5\frac{N}{C}$ away from the charged particle

$6\cdot 10^5\frac{N}{C}$ towards the charged particle

$6\frac{N}{C}$ away from the charged particle

$6\cdot 10^8\frac{N}{C}$ towards the charged particle

$6\cdot 10^3\frac{N}{C}$ away from the charged particle

Explanation

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.

Potential problem

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

to the right of

to the left from

to the right of

to the left of

There is no point on the line where the electric potential is zero

Explanation

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.

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.

None of these

Explanation

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.

Physics2set1q8

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

Points A and B are point charges.

$22.5 \frac{N}{C}$ in the direction

$8.99 \frac{N}{C}$ in the direction

$5.24 \frac{N}{C}$ in the direction

Explanation

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.

Electric field problem

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

to the right of

to the left of

to the right of

There is no point where the electric field is zero

Explanation

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:

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

Coloumbs law for varsity

If charge has a value of $-9*10^{-13}C$ , charge has a value of $8*10^{-13}C$ , and is equal to , what will be the magnitude of the force experienced by charge ?

$|F|=2.59*10^{-14}N$

None of these

$|F|=5.55*10^{-14}N$

$|F|=1.11*10^{-14}N$

$|F|=6.75*10^{-14}N$

Explanation

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

Converting into

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

Plugging values into coulombs law

$F=9*10^{9}\frac{-9*10^{-13}8*10^{-13}}{.50^2}$

$F=-2.59*10^{-14}N$

Magnitude will be the absolute value

$|F|=2.59*10^{-14}N$

Coloumbs law for varsity

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

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

None of these

$|\overrightarrow{E}|=6.9*10^6\frac{N}{C}$

$|\overrightarrow{E}|=9.5*10^6\frac{N}{C}$

$|\overrightarrow{E}|=2.21*10^6\frac{N}{C}$

Explanation

Using the electric field equation:

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

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-8*10^{-9}}{.003^2}$

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

Magnitude is equivalent to absolute value:

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

Coloumbs law for varsity

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 ?

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

None of these

$|\overrightarrow{E}|=3.75*10^6\frac{N}{C}$

$|\overrightarrow{E}|=3.98*10^6\frac{N}{C}$

$|\overrightarrow{E}|=6.11*10^6\frac{N}{C}$

Explanation

Using the electric field equation:

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

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{19*10^{-9}}{.0045^2}$

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

Magnitude is equivalent to absolute value:

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

Coloumbs law for varsity

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

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

None of these

$|\overrightarrow{E}|=3.77*10^5\frac{N}{C}$

$|\overrightarrow{E}|=2.25*10^5\frac{N}{C}$

$|\overrightarrow{E}|=1.73*10^5\frac{N}{C}$

Explanation

Use the electric field equation:

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

Where is $9.0*10^9 N\frac{m^2}{C^2}$

is the charge, in

is the distance, in .

Convert to and plug in values:

$\overrightarrow{E}=9.0*10^9\frac{-48*10^{-9}}{.025^2}$

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

Magnitude is equivalent to absolute value:

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

Electric Fields

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