Question 1

Many devices, such as mobile phones and drones, use Li-ion batteries as storage for electric energy. For a particular battery used for drone flights, the maximum charge that can be stored in one full charge is listed as 6000 mAh (milli-ampere hour). The operating voltage is given as 15.2 V.

i. Determine this maximum charge in Coulomb (C).

ii. In the quadcopter shown, if the four motors are drawing total energy at a constant rate of 120 W at the operating voltage given above, determine the current that is drawn from the battery. Determine how long (in minutes) a battery at full charge can last if it is operating at this current.

iii. The air mass $m$ passing a motor propeller blade each second is given by

where the density of air ${\rho }_{\text{air}}=1.22{\text{kg m}}^{-3}$, the area $A$ swept out by a propeller blade here is $0.045{\text{m}}^2$ and $v_{\text{air}}$ is the downward velocity of air due to the rotating propeller blade. The kinetic energy of the air per second generated due to one propeller blade is thus

when the quadcopter is just hovering. If a motor is drawing $30\text{W}$ from the battery and $80\mathrm{\%}$ of the energy is transferred to kinetic energy of air, determine $v_{\text{air}}$ (in m/s). Also determine the upward force (in N) acting on each propeller blade.

Solutions

The problem involves calculating charge, current, power, flight time, and the forces/energy related to the drone's propellers. Let's tackle these one by one.

i. Maximum Charge in Coulombs

The maximum charge is given as $6000\text{mAh}$ (milli-ampere hour). To convert this into the standard SI unit for charge, the Coulomb ($\text{C}$), we need to convert $\text{mAh}$ to $\text{A}\cdot \text{s}$ (Ampere-second), since $\mathbf{1}\text{A}\cdot \text{s}\mathbf{=}\mathbf{1}\text{C}$.

Step 1: Convert milli-ampere ($\text{mA}$) to ampere ($\text{A}$)

Step 2: Convert hour ($\text{h}$) to second ($\text{s}$)

Since there are $3600\text{seconds}$ in $1\text{hour}$:

The maximum charge stored in the battery is $\mathbf{21600}\text{C}$.

ii. Current, Power, and Flight Time

Here, we're using the relationships between Power ($P$), Voltage ($V$), and Current ($I$), and then using the charge capacity to find the duration.

1. Determine the Current Drawn

The total power drawn by the four motors is given as $P=120\text{W}$, and the operating voltage is $V=15.2\text{V}$. The relationship is:

We can solve for the current ($I$):

The total current drawn from the battery is approximately $\mathbf{7.89}\text{A}$ (rounded to three significant figures).

2. Determine the Battery Duration

The fundamental relationship between Charge ($Q$), Current ($I$), and Time ($t$) is:

We'll use the charge in Ampere-hours ($\text{Ah}$) and the current in Amperes ($\text{A}$) to get the time in hours ($\text{h}$), and then convert to minutes.

• Charge capacity: $Q=6.000\text{Ah}$
• Current: $I\approx 7.8947\text{A}$

3. Convert Time to Minutes

Since there are $60\text{minutes}$ in $1\text{hour}$:

The battery can last for approximately $\mathbf{45.6}\text{minutes}$ at this constant current draw.

iii. Air Velocity and Upward Force

This part delves into the mechanical power and thrust generated by a single propeller. When the drone is hovering, the total upward force (thrust) from the four propellers must equal the weight of the drone.

1. Determine $v_{\text{air}}$ (Downward Air Velocity)

The motor draws $30\text{W}$, and $80\mathrm{\%}$ of this power is transferred to the kinetic energy of the air. This means the mechanical power ($P$) generated by one propeller is:

The problem provides the formula for this kinetic energy per second (Power $P$):

We can now solve for $v_{\text{air}}$:

Plugging in the known values:

• $P_{\text{mech}}=24\text{W}$
• ${\rho }_{\text{air}}=1.22{\text{kg m}}^{-3}$
• $A=0.045{\text{m}}^2$

Taking the cube root:

The downward velocity of the air due to one propeller is approximately $\mathbf{9.56}\text{m/s}$.

2. Determine the Upward Force (Thrust)

The upward force, or Thrust ($F$), on the propeller is found using the momentum principle. The force is equal to the rate of change of momentum of the air:

Since the velocity $v_{\text{air}}$ is constant for hovering, this simplifies to:

We have the expression for $\frac{dm}{dt}$ from the problem: $\frac{dm}{dt}={\rho }_{\text{air}}A{v}_{\text{air}}$.

Plugging in the known values:

The upward force (thrust) acting on each propeller blade is approximately $\mathbf{5.02}\text{N}$ (rounded to three significant figures).

Question 2

A dart with a sticky nose has a total mass of $m=15.0\text{g}$ placed vertically on the ground. The sticky-nose dart faces upward on the spring and pushes the spring downward until the coils all press together, as in the figure given. The spring has a natural length of $4.30\text{cm}$. The compressed spring has a length of $1.40\text{cm}$. The spring constant is $29.4\text{N/m}$.

i. Determine the maximum height (in m) of the dart with sticky nose from the ground. You can ignore the effects of air resistance and assume that the spring is massless.
ii. Provide one suggestion so that the answer in part (i) will be larger.
iii. If air resistance $F_R$ on the dart with sticky nose is not negligible and is proportional to its velocity $v$, sketch two arrows, labelled with $F_R$: one arrow for when the dart is moving up and another arrow for $F_R$ when the dart is moving down.

iv. The graph below indicates how the vertical velocity of the dart will change with time just after it loses contact with the spring if air resistance is negligible. Without doing any calculations, sketch on the same axes to show how the velocity will vary if air resistance is not negligible. Label your line “N”.

Solutions

This is a classic problem where potential energy stored in the spring is converted into kinetic energy, and then into gravitational potential energy.

Part i: Maximum Height of the Dart

We can solve this problem by applying the principle of conservation of mechanical energy. We'll consider the system at two key points:

1. Initial State (A): The dart is at rest on the fully compressed spring.
2. Final State (B): The dart is at its maximum height and is momentarily at rest.

The total mechanical energy $E$ of the system (spring + dart + Earth) is conserved:

The energy at any point consists of:

• Elastic Potential Energy (${\text{PE}}_{\text{el}}$): Stored in the spring, $\frac{1}{2}k{x}^2$.
• Gravitational Potential Energy (${\text{PE}}_{\text{grav}}$): Due to height, $mgh$.
• Kinetic Energy ($\text{KE}$): Due to motion, $\frac{1}{2}m{v}^2$.

1. Identify Given Parameters and Calculate Spring Compression

First, let's gather and convert our given values to standard SI units (meters and kilograms):

• Mass of the dart, $m=15.0\text{g}=15.0\times {10}^{-3}\text{kg}=0.0150\text{kg}$
• Spring constant, $k=29.4\text{N/m}$
• Natural length of the spring, $L=4.30\text{cm}=0.0430\text{m}$
• Compressed length of the spring, $\ell =1.40\text{cm}=0.0140\text{m}$

The compression distance $x$ of the spring is the difference between its natural and compressed lengths:

2. Energy at the Initial State (A)

Let's set the lowest point of the dart (when the spring is fully compressed) as the reference level for gravitational potential energy, so $h_A=0\text{m}$.

• Kinetic Energy (${\text{KE}}_A$): The dart is initially at rest, so ${\text{KE}}_A=0$.
• Gravitational Potential Energy (${\text{PE}}_{\text{grav},A}$): At the reference level, ${\text{PE}}_{\text{grav},A}=mg{h}_A=0$.
• Elastic Potential Energy (${\text{PE}}_{\text{el},A}$): The spring is compressed by $x$.

The total initial energy is:

3. Energy at the Final State (B)

At the maximum height $H$ (measured from the ground), the dart momentarily stops.

• Kinetic Energy (${\text{KE}}_B$): The dart is momentarily at rest at its maximum height, so ${\text{KE}}_B=0$.
• Elastic Potential Energy (${\text{PE}}_{\text{el},B}$): The spring is now fully uncompressed (at its natural length), so ${\text{PE}}_{\text{el},B}=0$.
• Gravitational Potential Energy (${\text{PE}}_{\text{grav},B}$): The dart is at the maximum height $H$.

The total final energy is:

4. Apply Conservation of Energy

Set the initial energy equal to the final energy:

Now, solve for the maximum height $H$:

5. Calculation

Substitute the values into the equation:

First, calculate the numerator (initial elastic potential energy):

Next, calculate the denominator ($2mg$):

Finally, calculate $H$:

Rounding to three significant figures (based on the given values):

The maximum height of the dart from the ground is $0.0421\text{m}$.

Part ii: Suggestion to Increase the Maximum Height

The equation for the maximum height $H$ is:

To increase the maximum height $H$, we need to either increase the numerator ($k$ or $x$) or decrease the denominator ($m$ or $g$). Since $g$ (acceleration due to gravity) is fixed on Earth, we look at the other parameters:

Suggestion: Decrease the mass ($m$) of the dart.

• A lighter dart will have less gravitational potential energy stored for the same amount of initial spring energy, allowing it to reach a greater height. This is an easy modification to the dart itself.

Other valid suggestions would be:

1. Increase the spring constant ($k$): Use a stiffer spring.
2. Increase the compression distance ($x$): Compress the spring further if possible (the current problem states it's pressed until the coils all press together).

Part iii: Sketching Air Resistance Arrows ($F_R$)

That's an excellent follow-up question! Bringing in air resistance (a non-conservative force) significantly changes the dynamics and requires us to think carefully about the forces acting on the dart.

The key principle here is that air resistance (or drag force) always opposes the direction of motion. Since the air resistance $F_R$ is proportional to velocity $v$ (i.e., $F_R\propto v$), it will act opposite to $v$.

Here is the sketch with the two arrows labeled $F_R$:

Explanation of the Arrows

1. Dart Moving Up (Ascending):
   • The dart is moving upward.
   • The air molecules are pushing against the sticky nose and body of the dart to slow it down.
   • Therefore, the air resistance force ($F_R$) is directed downward, acting against the velocity.
2. Dart Moving Down (Descending):
   • The dart is moving downward (after reaching its maximum height).
   • The air molecules are again pushing against the dart to slow it down.
   • Therefore, the air resistance force ($F_R$) is directed upward, acting against the velocity.

Impact on Maximum Height

Just a quick side note: If we included this non-negligible air resistance in the energy calculation from part (i), the maximum height would be less than $0.0421\text{m}$. Why? Because the air resistance force does negative work on the dart throughout its flight, taking energy out of the mechanical system and converting it into heat, thus reducing the total energy available to become gravitational potential energy.

Part iv: Velocity-Time Graph with Air Resistance

That's an excellent conceptual question! It challenges us to think about how a non-conservative force like air resistance alters the idealized motion we typically study first.

The straight line on the graph represents the motion under negligible air resistance. This motion is governed solely by gravity, resulting in a constant downward acceleration of $a=-g$ (assuming up is positive), which gives the line its constant negative slope.

When air resistance ($F_R$) is not negligible, the motion changes significantly. Remember that the net force, and therefore the acceleration, is no longer constant.

Here is a sketch of the velocity-time graph with the new line labeled "N":

Key Features of the New Line (N)

The line "N" must show a continuously changing (non-constant) slope, which means the line will be a curve instead of a straight line.

1. Motion Upward ($v>0$):

   • The forces acting on the dart are gravity ($mg$, down) and air resistance ($F_R$, down).
   • The net downward force is $F_{\text{net}}=mg+F_R$.
   • Since $F_R$ is a downward force, the total downward force is greater than just gravity.
   • Therefore, the downward acceleration ($a=F_{\text{net}}/m$) is greater than $g$ (the slope is more negative).
   • As the dart slows down ($v$ decreases), $F_R$ decreases, so the magnitude of the slope also decreases slightly (becomes less negative) until the peak. The line starts steeper than the straight line.

2. At Maximum Height ($v=0$):

   • Air resistance $F_R$ is proportional to velocity $v$, so $F_R=0$ when $v=0$.
   • The acceleration is momentarily $a=-g$.
   • The curve must cross the horizontal axis with the same slope as the ideal straight line.

3. Motion Downward ($v<0$):

   • The forces acting on the dart are gravity ($mg$, down) and air resistance ($F_R$, up).
   • The net downward force is $F_{\text{net}}=mg-F_R$.
   • Since $F_R$ is an upward force, the total downward force is less than just gravity.
   • Therefore, the downward acceleration is less than $g$ (the slope is less negative).
   • The curve is flatter than the straight line. As the dart accelerates downward, $v$ increases (becomes more negative), $F_R$ increases, and the slope continues to decrease in magnitude, leveling off toward a terminal velocity (where $F_R=mg$).

In summary, the curve "N" starts steeper, momentarily matches the slope of the straight line at $v=0$, and then becomes flatter as it falls.

Question 3

As shown in the figure below, two charges $-Q$ and $+3Q$ are fixed at two points $M$ and $N$ respectively in the $x$-$y$ coordinate system. The two points $M$ and $N$ are at the same distance $a$ from the origin.

i. Determine the electric field (in vector form) at point $P$ if $Q=8\mu \text{C}$ and $a=2\text{cm}$.

ii. If a charge $q=-8\mu \text{C}$ is placed at point $P$, calculate the potential energy of charge $q$ if $Q=8\mu \text{C}$ and $a=2\text{cm}$.

iii. Consider the two charges and the Gaussian surfaces (sphere of radius $6\text{cm}$ in both cases) below. In case A, both charges are at the center of the sphere. In case B, both charges are just outside the sphere as shown. Determine the total electric flux through the Gaussian surfaces in each case if $Q=8\mu \text{C}$?

Solutions

Before we dive in, let's quickly review the setup from the figure:

• Charge 1: $Q_M=-Q$ at point $M$ with coordinates $(-a,0)$.
• Charge 2: $Q_N=+3Q$ at point $N$ with coordinates $(+a,0)$.
• The observation point is $P$ with coordinates $(0,2a)$.

The constants we'll need are: $Q=8\mu \text{C}=8\times {10}^{-6}\text{C}$, $a=2\text{cm}=0.02\text{m}$, and Coulomb's constant $k=9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$.

i. Determine the Electric Field (Vector Form) at Point $P$

The electric field $\overrightarrow{E}$ at point $P$ is the vector sum of the electric fields created by each individual charge, $Q_M$ and $Q_N$. That is, ${\overrightarrow{E}}_P={\overrightarrow{E}}_M+{\overrightarrow{E}}_N$.

The formula for the electric field due to a point charge is:

where $r$ is the distance from the charge to the point. The direction is away from a positive charge and towards a negative charge.

Step 1: Determine the distance $r$ from each charge to $P$

The coordinates are $M(-a,0)$, $N(a,0)$, and $P(0,2a)$.

The distance from $M$ to $P$ is $r_M$:

The distance from $N$ to $P$ is $r_N$:

The distances are the same! $r_M=r_N=a\sqrt{5}$.

Step 2: Calculate the magnitude of ${\overrightarrow{E}}_M$ and ${\overrightarrow{E}}_N$

Let's use the given values: $a=0.02\text{m}$ and $Q=8\times {10}^{-6}\text{C}$.

Magnitude of ${\overrightarrow{E}}_M$ (due to $Q_M=-Q$):

Magnitude of ${\overrightarrow{E}}_N$ (due to $Q_N=+3Q$):

Step 3: Determine the vector components

The electric fields ${\overrightarrow{E}}_M$ and ${\overrightarrow{E}}_N$ point along the lines connecting $M$ to $P$ and $N$ to $P$, respectively. ${\overrightarrow{E}}_M$ is towards $M$ (since $Q_M$ is negative), and ${\overrightarrow{E}}_N$ is away from $N$ (since $Q_N$ is positive).

Let $\theta$ be the angle that the line $MP$ or $NP$ makes with the horizontal ($x$-axis).

From the geometry:

Vector ${\overrightarrow{E}}_M$ (towards M):

The vector points in the $-x$ direction and $-y$ direction.

Vector ${\overrightarrow{E}}_N$ (away from N):

The vector points in the $-x$ direction and $+y$ direction.

Step 4: Sum the components to find ${\overrightarrow{E}}_P$

$x$-component ($E_x$):

$y$-component ($E_y$):

The Electric Field in Vector Form:

Alternatively, using the numerical approximation:

ii. Calculate the Potential Energy of Charge $q$ at Point $P$

The electric potential energy ($U$) of a charge $q$ placed at a point $P$ is given by:

where $V_P$ is the electric potential at point $P$ due to the source charges ($Q_M$ and $Q_N$).

Step 1: Calculate the Electric Potential $V_P$

The potential $V_P$ is the scalar sum of the potentials created by each charge. The formula for the potential due to a point charge is:

Note that charge signs are included in the potential calculation, and the result is a scalar.

Since $r_M=r_N=r=a\sqrt{5}$, and $Q_M=-Q$, $Q_N=+3Q$:

Plugging in the values: $k=9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$, $Q=8\times {10}^{-6}\text{C}$, $r=a\sqrt{5}\approx 0.0447\text{m}$.

Using the exact form:

Step 2: Calculate the Potential Energy $U$

The charge being placed at $P$ is $q=-8\mu \text{C}=-8\times {10}^{-6}\text{C}$.

The Potential Energy is:

Alternatively, using the numerical approximation:

iii. Determine the total electric flux through the Gaussian surfaces

The Underlying Principle: Gauss's Law

Gauss's Law states that the total electric flux (${\mathrm{\Phi }}_E$) through any closed surface (a Gaussian surface) is directly proportional to the net electric charge ($Q_{\text{enclosed}}$) enclosed by that surface.

The mathematical expression is:

Where ${ϵ}_0$ is the permittivity of free space (${ϵ}_0\approx 8.85\times {10}^{-12}{\text{C}}^2/(\text{N}\cdot {\text{m}}^2)$).

The key takeaway is that the size and shape of the Gaussian surface, and the exact positions of the charges outside the surface, do not affect the total flux. Only the net charge inside the surface matters.

The given value is $Q=8\mu \text{C}=8\times {10}^{-6}\text{C}$.

Case A: Both Charges are at the Center of the Sphere

In Case A (left image), the Gaussian sphere encloses both the charge $-Q$ and the charge $+3Q$.

Step 1: Calculate the Net Enclosed Charge ($Q_{\text{enclosed}}$)

Substituting the value of $Q$:

Step 2: Calculate the Total Electric Flux (${\mathrm{\Phi }}_{E,A}$)

Using Gauss's Law:

Case B: Both Charges are Just Outside the Sphere

In Case B (right image), both the charge $-Q$ and the charge $+3Q$ are outside the dashed Gaussian sphere.

Step 1: Calculate the Net Enclosed Charge ($Q_{\text{enclosed}}$)

Since both charges are outside the closed Gaussian surface, the net charge enclosed by the surface is zero.

Step 2: Calculate the Total Electric Flux (${\mathrm{\Phi }}_{E,B}$)

Using Gauss's Law:

The total electric flux through the surface is zero. (Note: The electric field $\overrightarrow{E}$ at the surface is not zero, but the net flux passing through the surface is zero—for every field line entering, one must exit).

Summary of Results

Note on Understanding the Concept

This problem beautifully illustrates the power of Gauss's Law. It tells us that we don't need to know the complex electric field pattern on the surface to find the total flux.

Think of electric flux like the amount of water flowing through a fishing net (your Gaussian surface).

• Case A: If you put a water source ($+3Q$) and a water sink ($-Q$) inside the net, the net flow through the surface is $3-1=2$ units out. The net enclosed charge is $2Q$.
• Case B: If the water source and sink are outside the net, the water field lines (electric field lines) must both enter and exit the net. For every line that enters, one eventually leaves, so the net flow through the net is zero. The net enclosed charge is $0$.

Do you feel confident with the distinction between the total electric field and the total electric flux, especially when charges are outside the Gaussian surface?