AP Physics 2 - Electric Potential Energy

Suppose that a point charge of 1 Coulomb undergoes a change in which it is moved from point A to point B while in the presence of an external electric field. During this transposition, it undergoes a voltage change of . What change in electrical potential energy occurs in this scenario?

Explanation

In order to solve for electrical potential energy, we'll need to remember the equation for it.

$U=k\frac{q_{s}q_{t}}{r}$

In the above expression, represents electrical potential energy, $q_{s}$ and $q_{t}$ represent different point charges, and represents the distance between their centers. In this example, one of these charges will be the source of the external electric field, while the other charge will be the one that is undergoing a transposition from point A to point B.

Furthermore, we can remember the equation for voltage:

$V=k\frac{q_{s}}{r}$

With both these equations in mind, we can combine the two:

$U=Vq_{t}$

This above expression tells us that the electrical potential energy of a system is directly proportional to the voltage change and to the charge of the test charge that is undergoing the voltage change.

Plug in the values and solve for electrical potential energy:

What is the electric potential between two terminals of a cell if it requires $\textup {7J}$ of work to transfer $\textup {3C}$ between the terminals?

$\frac{7}{3}\textup{V}$

$\frac{3}{7}\textup{V}$

$\frac{7}{3}\textup{A}$

$\frac{3}{49}\textup{A}$

$\frac{3}{7}\textup{eV}$

Explanation

Write the formula for the potential difference.

$V=\frac{W}{Q}$

Substitute the work and charge. The unit is in volts.

$V=\frac{W}{Q}= \frac{7\textup{J}}{3\textup{C}}=\frac{7}{3}\textup{V}$

$6.1 \cdot 10^{20}$ electrons pass through a $20\Omega$ resistor in 11 minutes. What is the potential difference across that resistor?

$e=1.602\cdot 10^{-19}$

Explanation

We need to find the voltage, and we have the resistance, so if we can find the current, then we can use Ohm's Law to find the voltage.

The definition of current is amount of charge that flows through a point in time, so current can be calculated using this equation:

$I=\frac{\Delta Q}{\Delta t}$

We're told how many electrons have passed through a resistor, and we know how much charge a single electron has. If we convert the number of electrons into total amount of charge, we can divide that number by the amount of time in seconds to find the current.

$6.1\cdot 10^{20} e^-\cdot 1.602\cdot 10^{-19}\frac{C}{electron}=97.722 C$

$11min\cdot60\frac{s}{min}=660s$

Now we have the amount of charge in and the amount of time in seconds, so we divide the two.

$I=\frac{\Delta Q}{\Delta t}$

$I=\frac{97.722}{660}$

Now that we have the current, we can use Ohm's Law to find the voltage.

Therefore, the potential difference is 2.96V.

In a region of space, there is an electric field whose magnitude is $E=20\frac{N}{C}$ pointing due North. An electron enters this field traveling due North with an initial velocity $v=2.4\cdot10^{4}\frac{m}{s}$ . It enters the field at point A, where the electric potential is 1.5V. As it travels 2cm in the field to point B, the potential changes to 0V. What will the electron's velocity be when it arrives at point B?

$m=9.11\cdot10^{-31}kg$

$q=-1.6\cdot10^{-19}C$

$3.63\cdot 10^{5}\frac{m}{s}$

$4.8\cdot 10^{5}\frac{m}{s}$

$1.6\cdot 10^{5}\frac{m}{s}$

$3.6\cdot 10^{4}\frac{m}{s}$

Cannot be determined

Explanation

This problem is solved using the work-kinetic energy theory: $\Sigma W=\Delta K$ . In an electric field, work is equal to charge times change in potential: $W=q\Delta V$

Since an electron has a negative charge, decreasing potential increases its kinetic energy, the opposite of what would happen to a proton with its positive charge. Combine these equations:

$q\Delta V=K_f-K_i$

$q\Delta V=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Plug in known values and solve

$1.6\times10^{-19}\cdot 1.5=\frac{1}{2}\cdot 9.11\cdot 10^{-31}v_f^{2}-\frac{1}{2}\cdot9.11\cdot10^{-31}\cdot(2.4\cdot10^{4})^{2}$

$v_f=3.63\cdot 10^{5}\frac{m}{s}$

There is an electric field of between two parallel plates of $3.5\cdot 10^{-5}\frac{N}{C}$ . The two plates are apart.

Find the electric potential energy of a particle of charge placed right at the surface of the positive plate.

$3.37\cdot 10^{-16}J$

None of these

$7\cdot10^{-16} J$

$2.33\cdot10^{-16} J$

$9.65\cdot 10^{-16} J$

Explanation

Use the equations for potential electric energy and electric potential:

$PE_{electric} = Vq$

Substitute.

$PE_{electric} = Edq$

Plug in known values and solve.

$PE_{electric}= (2.75\cdot 10^{-9}C)(35\cdot 10^{-6}\frac{N}{C})(3.5\cdot 10^{-3}m)$

$PE_{electric} = 337\cdot 10^{-18} Nm = 337\cdot10^{-18} J=3.37\cdot 10^{-16}J$

A proton is placed into an electric potential of , determine the final velocity after it is released.

$.438\frac{m}{s}$

None of these

$.756\frac{m}{s}$

$2.54\frac{m}{s}$

$1.51\frac{m}{s}$

Explanation

Conservation of energy:

Assuming no external work done

Electric potential energy:

Solving for velocity:

$v=\sqrt{\frac{Vq}{.5m}}$

Plugging in values:

$v=\sqrt{\frac{1*10^{-9}*1.6*10^{-19}}{.5*1.67*10^{-27}}}$

$v=.438\frac{m}{s}$

Determine the electric potential energy of a $50\textup{ nC}$ charge in an electric potential of $60\textup{ V}$ .

$3.0*10^{-6} \textup{ J}$

None of these

$4.5*10^{-6}\textup{ J}$

$1.5*10^{-6}\textup{ J}$

$6.0*10^{-6}\textup{ J}$

Explanation

Converting units:

$50nC=50*10^{-9}C$

$60V=60\frac{N*m}{C}$

Use the equation for electric potential energy:

$PE_{elec}=V*q$

$PE_{elec}=60*50*10^{-9}$

$PE_{elec}=3.0*10^{-6}\textup{ J}$

A car is traveling at $300\frac{km}{hr}$ . It is carrying a charge against an electric field of $50\frac{N}{C}$ . Determine how far the car will travel before stopping.

$1.127*10^{14}m$

None of these

$3.31*10^{6}m$

$4.88*10^{14}m$

$2.11*10^{14}m$

Explanation

Force in an electric field:

$F_{Electric Field}=Eq$

Using conservation of energy, assuming no external work on system:

Definition of electric potential energy:

Definition of electric potential:

Combining equations:

Assuming final velocity is zero:

is the distance traveled

Converting $\frac{km}{hr}$ to $\frac{m}{s}$

$300\frac{km}{hr}*\frac{1 hr}{3600s}*\frac{1000m}{1 km}=83.3\frac{m}{s}$

Plugging in values:

$.5*1625*83.3^2_i=50*1*10^{-9}*(d_f-d_i)$

Solving for

$(d_f-d_i)=1.127*10^{14}m$

A ball of mass with missing electrons is accelerated with a electric field. Determine the final velocity.

$1.95*10^{-5}\frac{m}{s}$

None of these

$1.95*10^{5}\frac{m}{s}$

$2.21*10^{5}\frac{m}{s}$

$4.11*10^{5}\frac{m}{s}$

Explanation

Force in an electric field:

$F_{Electric Field}=Eq$

Using conservation of energy, assuming no external work on system:

Definition of electric potential energy:

Combining equations:

Assuming initial velocity and final electric potential is zero:

The charge, will be equal to the electron charge time the number of electrons missing,

Converting to and plugging in values:

$.5*.0019*v^2_f=1.5*1*10^{6}*1.6*10^{-19}$

$v_f=1.95*10^{-5}\frac{m}{s}$

A student is working on a laboratory exercise in which she measures the electric potential (voltage) at several points in an electric field. Before graphing the potential isolines, she measures the electric potential to be 3V at one point in the field and 7V at a point 3cm from the first point. What is the average electric field strength between the two points the student measured?

$133\frac{N}{C}$