AP Physics 2 - Electric Force Between Point Charges

How will the force between two positively charged objects change as they are brought closer together?

The repulsive forces will increase

The attractive forces will increase

The force will change from repulsive to attractive

The repulsive force will decrease

It is impossible to determine

Explanation

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 acceleration of B due to A.

$3.2*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

None of these

$6.6*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

$1.1*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

$8.4*10^{-4}\ \frac{\textup{m}}{\textup{ s}^2}$

Explanation

Using Coulomb's law:

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

Using

Combining equations:

$a=k\frac{q_1q_2}{mr^2}$

Converting to and to and plugging in values:

$a=9*10^9\frac{2*10^{-9}*4*10^{-9}}{10*10^{-3}.15^2}$

$a=3.2*10^{-4}\frac{m}{s^2}$

Two electrons are deep in space and apart. Determine the force of one electron on the other.

$2.3*10^{-28}N$

None of these

$7.5*10^{-18}N$

Explanation

Using

$F_{elec}=k\frac{q_1q_2}{r^2}$

Plugging in values:

$F_{elec}=9.0*10^9\frac{({-1.6*10^{-19}})^2}{1^2}$

$F_{elec}=2.3*10^{-28}N$

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?

There is no way to determine the new force

Explanation

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:

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

Coloumbs law for varsity

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

$|\overrightarrow{F}|=5.35*10^{-3}N$

None of these

$|\overrightarrow{F}|=2.59*10^{-3}N$

$|\overrightarrow{F}|=9.87*10^{-3}N$

$|\overrightarrow{F}|=1.89*10^{-3}N$

Explanation

Using the electric field equation:

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

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

is charge , in

is the distance, in .

Convert to and plug in values:

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

$\overrightarrow{F}=-5.35*10^{-3}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=5.35*10^{-3}N$

An electron is away from a point charge. It experiences a force of $2.6*10^{-16}N$ towards the point charge. Determine the value of the point charge.

$4.5*10^{-8}C$

None of these

$4.5*10^{-1}C$

$9*10^{-8}C$

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

Explanation

Using

$F=k\frac{q_1q_e}{r^2}$

Solving for

$F\frac{r^2}{k*q_e}=q_1$

Converting to and plugging in values

$-2.6*10^{-16}\frac{.500^2}{(9*10^{9})*-1.6*10^{-19}}=q_1$

*Note: a negative sign is used for the force because it is an attractive force, if it was a repulsive force, the opposite sign would be used.

$q=4.5*10^{-8}C$

What is the force experienced by a $20\mu C$ point charge away from a $30\mu C$ point charge?

The point charge experiences no force

Explanation

The equation to find the force from two point charges is called Coulomb's Law.

$F=\frac{kq_1q_2}{r^2}$

In this equation, is force in Newtons, is the respective charge value in $\mu C$ , is radius in meters, and is the Coulomb constant, which has a value of $9 \cdot 10^9 \frac{N*m^2}{C^2}$ .

Now, we just plug in the numbers. Note: the charge values are in microcoulombs (the Greek letter $\mu$ , called "mu," stands for micro), which is equal to $10^{-6}$ .

$F&=\frac{kq_1q_2}{r^2}$

$F=\frac{(9\cdot 10^9\frac{N*m^2}{C^2}) \cdot (20\cdot 10^{-6}\mu C) \cdot (30 \cdot 10^{-6}\mu C)}{(5m)^2}$

Therefore, the force experienced on either charge is 0.216N of force.

Coloumbs law for varsity

Charge has a charge of

The distance between their centers, is .

What is the magnitude of the force on due to ?

$|\overrightarrow{F}|=2.4*10^{-1}{N}$

None of these

$|\overrightarrow{F}|=4.5*10^{-1}{N}$

$|\overrightarrow{F}|=7.9*10^{-1}{N}$

$|\overrightarrow{F}|=9.1*10^{-1}{N}$

Explanation

Use Coulomb's law:

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

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

is charge , in Coulombs

is the distance, in meters.

Convert to and plug in values:

$\overrightarrow{F}=9.0*10^9\frac{19*10^{-9}*-28*10^{-9}}{.0045^2}$

$\overrightarrow{F}=-2.4*10^{-1}{N}$

Magnitude is equivalent to absolute value:

$|\overrightarrow{F}|=2.4*10^{-1}{N}$

Two charges are a fixed distance apart. Both charges have charge . If another charge of charge and mass is placed a distance $\frac{d}{4}$ from one of the charges and $\frac{3d}{4}$ from the other, what will be the magnitude of its acceleration the moment it's released?

$\frac{128kqQ}{9md^2}$

$\frac{16kqQ}{9md^2}$

$\frac{16kqQ}{md^2}$

$\frac{8kqQ}{3md}$

$\frac{8kqQ}{9md}$

Explanation

All we have to do is find the sum of the forces on the charge and divide by its mass. To find the force from each charge, we can use Coulomb's law:

$F=\frac{kq_{1}q_{2}}{d^2}$

Let's let the force from the charge a distance $\frac{d}{4}$ away be positive. That force is $\frac{kqQ}{(\frac{d}{4})^2}=\frac{16kqQ}{d^2}$ . The other force will be negative because it's acting in the opposite direction. This force is $\frac{-kqQ}{(\frac{3d}{4})^2}=\frac{-16kqQ}{9d^2}$ . Adding these two together we get