# How a Capacitor Discharge

### How does a capacitor work?

When a capacitor is connected across a power source, it starts charging up as current flows in the circuit.

Initially when the capacitor has not stored any charge, its potential difference is 0 V. As current flows, charges start storing across its plates, and the potential difference increases. The current also starts to decrease because the capacitor reduces the overall e.m.f of the circuit. When the potential difference of the capacitor equals to the e.m.f. of the power source, current stops flowing.

After the capacitor is fully charged, the switches are set to the figure above. The capacitor starts discharging and a current flows in the left part of the circuit. The potential difference across the capacitor follows an exponential curve.

### Wolfram Demonstration

Click this Wolfram Demonstration Project link to access the simulation. This simulation shows how the potential difference across the capacitor decreases during a discharge. The rate of discharge is dependent on the capacitance of the capacitor. A larger capacitor stores more charge. Although the maximum voltage across it is dependent on the power source, a large capacitor produces a larger current.

Procedure:

1. Adjust all the controls to the left such that d = 0.4, L = 1.5 and time = 0.
1. Observe that the capacitor discharges almost all its stored charge after about 50 s.
2. The capacitor has a capacitance of $0.481\mu \text{F}$ and its maximum stored charge is $21.6 \mu \text{C}$
2. Increase the separation d to 0.7.
1. Observe that the capacitance is now $0.275\mu \text{F}$ and maximum charge $12.4 \mu \text{C}$. This means that increasing the distance between the plates decreases the capacitance, and correspondingly the charge stored, since $Q=CV$ .
2. Lesser stored charge means that the current stops flowing after about 30 s.
3. Increase the length of side to 3.0.
1. Observe that the capacitance is now $1.1 \mu \text{F}$ and that the charge stored is now $49.4 \mu \text{C}$ . Increasing the cross-sectional area increases the capacitance and the stored charge.
2. Since the stored charge is larger, the capacitor takes a longer time to discharge to 0 V.

### Summary

1. Capacitance depends on the physical property. The capacitance is larger is the distance between the plates is small and the cross-sectional is large.
2. A large capacitor stores a larger charge, and takes a longer time to discharge. Conversely, it should take a larger time to charge a large capacitor to its maximum voltage.

# Capacitance

### What is a capacitor?

A capacitor is an electrical component that stores charge. It is usually made by having two parallel plates thinly separated by an insulating material. This material is also known as a dielectric. Current flows through an insulator if the voltage across it is sufficient. In a capacitor, the conducting plates are close enough for the current to flow, even through it is separated by the insulating dielectric. However, because of the dielectric, some charges are stored on the plates as the current flows, resulting in higher potential difference as time passes. When the potential difference across the capacitor equals to the e.m.f of the circuit, current stops flowing.

### Physics of capacitance

Capacitance is defined as

Capacitance is the ratio of the charge stored to potential difference across a capacitor.

Mathematically,

\begin{aligned}C=\frac{Q}{V} \end{aligned}

The SI  unit of capacitance is farad (F).

A capacitor of 1 farad has a stored charge of 1 C when the potential difference across it is 1 V.

### Combined Capacitance

Capacitors can be connected either in series or parallel.

In a series connection, the combined capacitance can be found from the relationship

\begin{aligned} V_\text{total} &= V_1 + V_2 + ...\\\frac{Q}{C_\text{total}} &= \frac{Q}{C_1}+\frac{Q}{C_2}+...\end{aligned}

Since the charge $Q$ stored in each capacitance is the same,

\begin{aligned}\frac{1}{C_\text{total}} &= \frac{1}{C_1}+\frac{1}{C_2}+...\end{aligned}

In a parallel connection, the combined resistance is derived as follow:

\begin{aligned} Q_\text{total}&=Q_1+Q_2+...\\C_\text{total}V&=C_1V+C_2V+...\\\end{aligned}

Since the potential difference $V$ is the same in a parallel circuit,

\begin{aligned} C_\text{total}=C_1+C_2+... \end{aligned}

## Summary

1. Capacitance is the ratio of the charge stored to the potential difference across the component.
2. SI unit of capacitance is farad (F).
3. The total capacitance of series connected capacitors is \begin{aligned} \frac{1}{C_\text{total}}=\frac{1}{C_1}+\frac{1}{C_2}+... \end{aligned}
4. The total capacitance of parallel connected capacitors is \begin{aligned} C_\text{total} = C_1+C_2+... \end{aligned}

## Review

### Question 1

The figure shows three capacitors connected in series with a cell of e.m.f. 3.0 V.

Calculate the p.d. across each capacitor.

### Solution

Since the charge stored in each capacitor is the same as the charge stored across all the capacitors, we shall go ahead and find the charge stored in the combined capacitors and then use it to calculate the potential difference across each capacitor.

\begin{aligned} \frac{1}{C_\text{total}} &= \frac{1}{100 \times 10^{-3}}+\frac{1}{200 \times10^{-3}} + \frac{1}{400 \times 10^{-3}}\\C_\text{total}&= 0.0571\text{ F}\\ Q_\text{total}&=3.0 \times 0.0571\\&=0.1713\text{ C}\end{aligned}

Hence,

\begin{aligned} V_1&=\frac{0.1713}{100 \times 10^{-3}}\\&=1.713\text{ V}\\V_2&=\frac{0.1713}{200 \times 10^{-3}}\\&=0.857\text{ V} \\V_3&=\frac{0.1713}{400 \times 10^{-3}}\\&=0.428\text{ V}\end{aligned}

You can observe that the sum of all the three p.d. is the e.m.f. of the cell.

# Gravitation and Circular Motion

### Why does a orbiting satellite not fall to the Earth?

A common misconception among students is that orbiting satellites do not experience gravity. It is argued that since these satellites are not falling to the Earth, they must experience no gravity. This thought is not true because the very action of orbiting around the Earth shows that the satellite is under the influence of Earth’s gravity. Then why is the satellite not falling to the Earth?

The below simulation explains why satellites are not falling back to the Earth. In fact, these satellites are actually in the process of falling to the Earth. The simulation explains clearly how this works.

Satellite orbiting the Earth

So the answer to the above question is that the satellite is in fact, falling to the Earth. However, the Earth’s surface curves away at the same rate the the satellite is falling towards Earth. Hence, the satellite never falls onto the Earth.

It is important to note that this can happen because the satellite possesses an initial velocity that is parallel to the Earth’s surface. Without sufficient velocity of this component, the satellite would fall towards Earth.

### Gravitational force and circular motion

Planets revolve around the Sun and the Moon revoles around the Earth. These orbits are generally approximately circular, and the centripetal force to pull them around the orbits is the gravitational force.

\begin{aligned} \frac{GMm}{r^2} &= \frac{mv^2}{r} \\ \frac{GM}{r} &= {v^2}\\v&=\sqrt{\frac{GM}{r}}\end{aligned}

We can see that the linear speed, hence the angular speed, of an orbiting body is independent of the mass of the body. It is only dependent on the mass of the body producing the gravitational field and the distance of the orbiting body from the centre of the main mass.

To find the relationship between the radius and the period of orbit,

\begin{aligned} (\frac{2\pi r}{T})^2 &= \frac{GM}{r}\\T^2 &= \frac{4\pi^2}{GM}r^3 \end{aligned}

Hence we can see that the square of the period is proportional to the cube of the radius of orbit. The is the Kepler’s third law.

### Summary

1. When an object orbits around another mass, the centripetal force is gravitational in nature.

2. The square of the period is proportional to the cube of the radius of orbit. This is also known as the Kepler’s third law.

### Review

#### Question 1

Two stars orbit each other in a time of 3.0 years. Calculate the angular speed \begin{aligned} \omega \end{aligned} for each star. Given that the ratio \begin{aligned} \frac{M_1}{M_2}=2.0 \end{aligned} , and the separation of the starts is $3.0 \times 10^{12} \text{ m}$ , calculate the radii \begin{aligned} R_1 \end{aligned} and \begin{aligned} R_2 \end{aligned} .

#### Solution:

Angular speed is the angular displacement per unit time. Hence,

\begin{aligned}\omega &= \frac{2 \pi}{3.0 \times 365 \times 24 \times 3600}\\&=6.6\times 10^{-8}\text{rad s}^{-1} \end{aligned}

The two stars must be orbiting with the same angular speed. Otherwise, there will be a point in time when the inner star catches up and the centre of mass would change position. The gravitational force between the two stars would then change, changing the centripetal force. This is not possible for a stable star system, hence the angular speeds of the two stars must be the same.

Equating the gravitational force to the centripetal force,

\begin{aligned} \frac{GM_1M_2}{(R_1 + R_2)^2 } &= M_2 R_2 \omega^2\\\frac{GM_1}{(R_1 + R_2)^2} &= R_2 \omega^2 \end{aligned}

Similarly,

\begin{aligned} \frac{GM_2}{(R_1 + R_2)^2} &= R_1 \omega^2 \\\frac{M_1}{M_2} &= \frac{R_2}{R_1}\end{aligned}

Since \begin{aligned} \frac{M_1}{M_2} = 2.0 \end{aligned} ,

\begin{aligned} \frac{R_2}{R_1}&=2.0\\R_2&=2.0R_1\\R_1+2R_1 &= 3.0 \times 10^{12}\\R_1&=1.0\times 10^{12}\text{ m}\\R_2 &= 2.0 \times 10^{12} \text{ m} \end{aligned}

# Gravitation – Potential and Potential Energy

## Gravitational Potential and Potential Energy

Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.

Gravitational potential is defined to be negative since gravity is attractive in nature. A negative potential means that no external work is needed to bring a unit mass from infinity to that point, since the gravity-producing mass would be doing the work to pull the unit mass to that point.

Mathematically,

\begin{aligned} \phi = -\frac{GM}{r} \end{aligned}

Note that potential is a scalar quantity. Hence, the potential at a point is simply the algebraic sum of the potential of the different masses at that point.

\begin{aligned} \phi_\text{sum} = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2} - ... \end{aligned}

Similarly, gravitational potential energy is defined as

Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.

\begin{aligned} \text{GPE} = -\frac{GMm}{r^2} \end{aligned}

You may find this concept similar to gravitational force and field strength. Both gravitational force and potential energy invloves the product of two masses \begin{aligned} Mm \end{aligned} while field strength and potential involves just the gravity-producing mass \begin{aligned} M \end{aligned}.

## Gravitational Potential vs Field Strength

It is easy to compare the relative values of potential and field strength because of the similar form of equations.

\begin{aligned} g &= -\frac{GM}{r^2} \\ \phi &= -\frac{GM}{r}\end{aligned}

## GPE = mgh

In many junior Physics text, gravitational potential energy is quoted with the formula
\begin{aligned} \text{GPE} = mgh \end{aligned}
This formula assumes that the change in height in insignificant compare to the radius of the Earth. This formula also calculates the change in the potential energy due to a change in position.

The formula \begin{aligned} F_G = \frac{GMm}{r} \end{aligned} calculates the actual amount of potential energy a mass possess due to its position. This formula does not calculate the change in potential energy. To calculate the change in potential energy,

\begin{aligned}\text{change in GPE} &= \frac{GMm}{r_1} - \frac{GMm}{r_2}\\\end{aligned}

If \begin{aligned} r_1 \approx r_2 \end{aligned} ,

\begin{aligned} \text{change in GPE} &= GMm ( \frac{1}{r_1} - \frac{1}{r_2} ) \\&= GMm (\frac{r_2-r_1}{r_1 r_2})\\&=gm(r_2-r_1)\\&=mgh\end{aligned}

From this, we have our old formula \begin{aligned} \text{GPE} = mgh \end{aligned}

### Summary

1. Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.
2. Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.
3. \begin{aligned} \phi = -\frac{GM}{r} \end{aligned}
4. The gravitational potential has a larger magnitude than the field strength.
5. When the change in height is small, we may use \begin{aligned} \text{change in GPE} = mgh \end{aligned}

# Gravitation – Force and Field Strength

## Concept of a Gravitational Field and Force

A gravitational field is a region in which a mass experiences a force.

A mass, \begin{aligned} m \end{aligned} that is present inside a gravitational field experiences a gravitational force. This gravitational field is produced by another mass \begin{aligned} M \end{aligned} .

The amount of force experienced by the mass \begin{aligned} m \end{aligned} is directly proportional to the product of the two masses and indirectly proportional the the square of the separation.

\begin{aligned} F_G=\frac{GM m}{r^2} \end{aligned}

The separation \begin{aligned} r \end{aligned} is the distance between the centre of mass of the two masses. We often assume that the two masses are point masses if the separation is large relative to the radius of the masses. If the separation is not large, then it is important to use the distance between the centre of mass of the two masses. We should not use the separation between the two surfaces of the masses. Furthermore, one may safely assume that for a uniform sphere, the centre of mass is the centre of the sphere.

There are two regions about the field produced by a mass: the region outside the mass and the region inside. The region outside the mass follows the inverse exponential relationship of 1/r^2.

Inside the mass, the relation is direct proportional to the distance from the centre of the mass. This is because as you proceed nearer to the centre of the mass, there is less mass “below” you. The part of the mass “above” you pulling you “up” is offset by the mass “below” you pulling you down.

### Gravitational Field Strength

Gravitational field strength is often misunderstood. Its definition is

Gravitational field strength at a point is the gravitational force acting on a unit mass at that point.

From the definition,

\begin{aligned} \text{force} &= m \times g\\mg &= \frac{GMm}{r^2}\\g &= \frac{GM}{r^2}\end{aligned}

We can observe that gravitational field strength only depends on the gravity-producing mass and the distance from it. It is not dependent on the test mass.

There is only one value of gravitational field strength at any particular point since we are always comparing the gravitational force on one unit mass. If there are multiple masses creating gravitational fields, the gravitational field strength at any particular point would be the vector sum of all the field strengths due to the the different masses.

### Field Strength on the surface of the Earth

When we calculate weight of an object, we always use the formula \begin{aligned} w = mg \end{aligned}

\begin{aligned} g \end{aligned} is referred to as the gravitational field strength (although it is commonly stated as the gravitational acceleration). Near the Earth’s surface, the field strength of the Earth is

\begin{aligned} g &=G\frac{M}{r_\text{Earth radius}^2}\\&=G\frac{M}{6400000^2}\end{aligned}

The field strength at 10 km above the surface would be \begin{aligned} G\frac{M}{6410000^2} \end{aligned}.

The difference among them is negligible. Hence we assume that the gravitational field strength on Earth’s surface is constant.

### Activity 1

Access the online PHET simulation.

Objective:
What is the relationship between the force acting on m2 by m1 and the force of m1 on m2?

1. Using the default values, observe the force on m2 by m1 and the force on m1 by m2.
What do you observed?
2. Change the mass of m1 to another value.
What do you observed?
3. Explain.

### Activity 2

Access the same simulation as activity 1. Also open this activity sheet to download a copy of the activity on iCloud.

Objective:
What is the relationship between the field strength with distance?

1. Change m2 to 1 kg, while m1 remains as 50 kg.
2. Drag m2 to the 10 m mark, and m1 to the 0 m mark. You may not be able to do this exactly, but an approximate position would be fine.
3. Record the force on m2 by m1 in the table.
4. Repeat this for distances 9, 8, 7, 6, 5, 4, 3, 2 and 1 m.
5. Observe the shape of the graph.

Question:

1. Will the gravitational field strength ever reaches zero?
2. At which point(from 0 m to infinity) is the field strength the strongest?

### Summary

1. understand the concept of a gravitational field as an example of a field of force and define gravitational field strength as force per unit mass.
2. understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.
3. recall and use Newton’s law of gravitation in the form \begin{aligned} F=G\frac{Mm}{r^2} \end{aligned}

### Review

##### Question 1

Geostationary satellites are satellites that orbit around the Earth with a period of 24 hours. The satellite would appear as a stationary point relative to an observer on Earth. Calculate the distance above the Earth’s surface of a geostationary satellite. Properties of Earth

##### Solution

Since the centripetal force of the satellite is due to gravity,

\begin{aligned} \frac{GMm}{r^2} &= mr\omega ^2\\\frac{GM}{r^2} &= r \omega ^2 \\r &= \sqrt[3]{\frac{GM}{w}} \end{aligned}

$\omega$ is known, since the geostationary satellite must make one orbit in one day,

\begin{aligned} \omega &= \frac{2 \pi}{24 \times 3600}\\&= 7.3\times 10^{-5} \text{ rad s}^{-1}\\r&=42.1\times 10^6\text{ m}\end{aligned}

Hence, the distance above Earth’s surface is

\begin{aligned} d &= (42.1 - 6.38) \times 10^6 \text{ m}\\ &= 35.7 \times 10^3 \text{ km} \end{aligned}

It is interesting to note that the distance of Moon from Earth is about $370 \times 10^3 \text{ km}$. Hence, a geostationary satellite is about 10% of the distance from Earth to the Moon.

# DEEP Learning – Singapore 30 April 2016

### iBooks Author – Design

Design is an important aspect of writing a book. A book that contains a lot of information in text does not catch the attention of its readers. Capturing this information into visuals is an important skill that an author can use to make his book more popular.

Graphic design is visual, which means That it involves our sense of sight. The skillful ways which graphics designers arrange and present words and images using text, colour, size and composition – makes it possible for us to see, remember (and enjoy!) complex ideas.

There are a few points that you can take note when you design an iBook:

1. Contrast
2. Repetition
3. Alignment
4. Proximity

### Contrast

Contrast can be obtained by the clever use of colours, shapes, size and positions.

### Repetition

Repetition refers to reusing common elements in your book. For example, assign colours to specific elements and be consistent throughout your book. Readers would be able to link the meaning of your content by the perception of colour.

### Alignment

Text blocks and image edges should be alignment to give the readers a neat and well organised content. It makes visual search for contents easy for the reader too.

### Proximity

COntents that are related should be closed to each other. For example, in the design of a travel brochure, all the facts about the history of the place should be together, while places to eat should then be grouped together, away from the first block.

## Colours

A colour wheel shows the primary, secondary and tertiary colours. Primary colours are yellow, red and blue. Secondary colours are orange, purple and green. Tertiary colours occupy between the primary and secondary colours.

To choose a set of colours that are compatible to a specific colour, the different colours should extend on the colour wheel from this colour until the next primary colour. In the example, if your chosen colour is blue, then a set of campatible colours should extend from either sides until, but not including, yellow and red.

If you have limited colours in your design, then you could use hues to represent more colours. In using hues, there are three schemes:

1. Analogous, which are the hues on either sides of your chosen colour,
2. Complement, which are the hues on the opposite sides of the colour wheel, and
3. Split complement, which are the two hues on either sides of the opposite hue.

## Summary

Positioning of text blocks and illustrations should follow the CRAP rule: Contrast, Repetition, Alignment and Proximity. Contrast can be brought about by the creative use of colours, shapes, size and positions. Repetition in reusing of elements to represent certain means should be adhered to, so that readers can draw a meaning from the patterns. Alignment makes the contents neat and easy to search for patterns. Proximity groups contents of similar nature together so that readers can obtain most of the information efficiently.

Colours make your iBook attractive. Choice of colours can be made by following the colour wheel. From your chosen colour, you may use any colours to the left and right until the next primary colour, or if you have limited colours, you may choose colour hues. When choosing hues, you may choose using the analogous, complementary or split complementary colour schemes.

# Work done in a Capacitor

Capacitor is an electrical component that stores charge. As charges are stored, potential energy in the capacitor also increases. In this post, I would like to explain how to calculate the energy stored in a capacitor.

##### Question

The switch is closed at time t = 0 s. During the time interval from t = 0.0 s to t = 4.0 s, 15 mC charge passes through the resistor.

Calculate the energy transferred by the battery during this 4.0 seconds and the energy stored in the capacitor after 4.0 s.

#### Definition of Potential Difference

In many textbooks, we learnt that the potential difference across a component is the energy released for every unit charge passing through it.

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

If we are to calculate the energy released by the battery, it would be

$W=V \times Q = 9.0 \times 15 \times 10^{-3} =0.135 \text{J}$

#### Energy stored in the capacitor

However, the energy stored in the capacitor cannot be 0.135 J. This is because a capacitor’s potential difference changes as it stores charge. The equation above requires that the p.d. remains constant. Hence, we need to use another formula.

Since capacitance is defined as the charge stored for every unit potential difference,

$Q=VC$

i.e. the charge stored is directly proportional to the potential difference across it. Since the work done is the area under a charge-voltage graph, we will use the formula for a triangle.

$W=\frac{1}{2}QV$

At time t = 4.0 s,

$V=\frac{15 \times 10^{-3}}{2000 \times 10^{-6}}=7.5 \text{V}$

Hence,

$W=\frac{1}{2} \times 15 \times 10^{-3} \times 7.5 = 0.05625 \text{J}$

#### Lost energy?

The energy stored in the capacitor(0.05625 J) is not the same as the energy transferred from the battery(0.135 J) because heat is lost through the resistor as current flows.

# IGCSE Definitions

This document contains the common definitions that may be asked in the Cambridge IGCSE examinations.

IGCSE Definitions

Radioactive decay is both random over space and time, and radioactive decay is spontaneous.

### Random

Random over space means that given a large number of radioactive nuclei, it would be impossible to predict which nuclei would decay next. Random over time means that you cannot predict when a nuclei would decay at any time.

Given a large number of nuclei, it can be safe to predict that approximately half of the original radioactive nuclei would decay after one half-life. However, the exact number of nuclei decay would not be identical given two samples of equal number of original radioactive nuclei.

# Ultrasound Calculations

The table provides the data for the acoustic impedance and absorption coefficients for muscle and bone.

A parallel beam of ultrasound of intensity $I_0$ enters the muscle of thickness 3.0 cm as shown. The ultrasound is then reflected at the tissue boundary and returns to the surface of the muscle.

Calculate the intensity, in terms of $I_0$, that is received when the ultrasound returns back to the surface of the muscle.

There are three important parts to solve this problem. The first part would be the attenuation of intensity inside the muscle. The second part would be the reflection of ultrasound at the muscle-bone boundary. The last part would be the intensity attenuation inside the muscle back to the surface.

$I=I_0 e^{-21 times 0.03}$