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# A-level Physics (Advancing Physics)/Electric Potential Energy/Worked Solutions

k = 8.99 x 10$^9$ Nm$^2$C$^{−2}$
1. Convert 5 x 10$^{−13}$ J to MeV.
${\frac {5\times 10^{-13}}{1.6\times 10^{-19}}}=3125000{\mbox{ eV}}=3.125{\mbox{ MeV}}$
2. Convert 0.9 GeV to J.
$0.9\times 10^{9}\times 1.6\times 10^{-19}=1.44\times 10^{-10}{\mbox{ J}}$
3. What is the potential energy of an electron at the negatively charged plate of a uniform electric field when the potential difference between the two plates is 100V?
$\epsilon _{elec}=V_{elec}q=100\times 1.6\times 10^{-19}=1.6\times 10^{-17}{\mbox{ J}}=0.1{\mbox{ keV}}$
4. What is the potential energy of a 2C charge 2 cm from a 0.5C charge?
$\epsilon _{elec}={\frac {kQq}{r}}={\frac {8.99\times 10^{9}\times 2\times 0.5}{0.02}}=4.5\times 10^{11}{\mbox{ J}}$
Remember, coulombs are big!
5. What is represented by the gradient of a graph of electric potential energy against distance from some charge?
Force exerted on the charge with that energy.
# A-level Physics (Advancing Physics)/Electric Field

Electric field E is the force per. unit charge caused by an electric field:
$E_{electric}={\frac {F_{electric}}{q}}$
The unit of electric field is NC$^{−1}$ or Vm$^{−1}$.In general, the electric field is the rate of change of electric potential (voltage) with respect to distance:
$E_{electric}=-{\frac {dV_{electric}}{dx}}$

## Special Cases

There are two different types of field which you need to know about. Uniform fields occur between two plates with opposite charges. Here, the electric field is simply:
$E_{electric}={\frac {V_{electric}}{d}}$
A charged sphere also has an electric field. To gain a formula for this, we divide the formula for force around a charged sphere by q, so:
$E_{electric}={\frac {Q}{4\pi \epsilon _{0}r^{2}}}$,
where ε$_0$ = 8.85 x 10$^{−12}$C$^2$N$^{−1}$m$^{−2}$.
The electric field around a point charge is called a radial field. The field strength is highest at the centre and decreases as the distance from the centre increases. This is reflected in the above formula, which shows that E$_{electric}$ is proportional to ${\frac {1}{r^{2}}}$.

## Field Lines

We can represent electric field using field lines. These go from positive charge to negative charge. They are more closely packed together when the electric field is stronger. In a uniform field, they look like the following:
Around two oppositely charged spheres (known as a dipole), they look like the following:

## Questions

1. Two metal plates are connected to a 9V battery with negligible internal resistance. If the plates are 10 cm apart, what is the electric field at either of the plates?
2. What is the electric field at the midpoint between the plates?
3. The charge on an electron is -1.6 x 10$^{−19}$ C. What is the electric field 1μm from a hydrogen nucleus?
4. What is the direction of this field?
5. A 2C charge is placed 1m from a -1C charge. At what point will the electric field be 0?
Worked Solutions
# A-level Physics (Advancing Physics)/Doppler Effect

The Doppler effect is a change in the frequency of a wave which occurs if one is in a different frame of reference from the emitter of the wave. Relative to us, we observe such a change if an emitter of a wave is moving relative to us.
All waves travels in a medium. So, they have a velocity relative to this medium v. They also have a velocity relative to their source v$_s$ and a velocity relative to the place where they are received v$_r$. The frequency at which they are received f is related to the frequency of transmission f$_0$ by the formula:
$f=\left({\frac {v+v_{r}}{v+v_{s}}}\right)f_{0}$
The Doppler effect can be used to measure the velocity at which a star is moving away from or towards us by comparing the wavelength received, λ, with the wavelength we would expect a star of that type to emit, λ$_0$. Since the speed of light c is constant regardless of reference medium:
$c=f\lambda =f_{0}\lambda _{0}$
Therefore:
$f={\frac {c}{\lambda }}$ and $f_{0}={\frac {c}{\lambda _{0}}}$
By substitution:
${\frac {c}{\lambda }}=\left({\frac {v+v_{r}}{v+v_{s}}}\right){\frac {c}{\lambda _{0}}}$
${\frac {1}{\lambda }}=\left({\frac {v+v_{r}}{v+v_{s}}}\right){\frac {1}{\lambda _{0}}}$
$\lambda ={\frac {\lambda _{0}(v+v_{s})}{v+v_{r}}}$
In this case, v is the speed of light, so v = c. Relative to us, we are stationary, so v$_r$ = 0. So:
$\lambda ={\frac {\lambda _{0}(c+v_{s})}{c}}$
${\frac {\lambda }{\lambda _{0}}}={\frac {(c+v_{s})}{c}}=1+{\frac {v_{s}}{c}}$
If we call the change in wavelength due to Doppler shift Δλ, we know that λ = λ$_0$ + Δλ. Therefore:
${\frac {\lambda _{0}+\Delta \lambda }{\lambda _{0}}}=1+{\frac {\Delta \lambda }{\lambda _{0}}}=1+{\frac {v_{s}}{c}}$
So, the important result you need to know is that:
${\frac {\Delta \lambda }{\lambda _{0}}}={\frac {v_{s}}{c}}=z$
This value is known as the red-shift of a star, denoted z. If z is positive, the star is moving away from us - the wavelength is shifted up towards the 'red' end of the electromagnetic spectrum. If z is negative, the star is moving towards us. This is known as blue shift. Note that we have assumed that v is much smaller than c. Otherwise, special relativity makes a significant difference to the formula.

## Questions

1. M31 (the Andromeda galaxy) is approaching us at about 120kms$^{−1}$. What is its red-shift?
2. Some light from M31 reaches us with a wavelength of 590 nm. What is its wavelength, relative to M31?
3. Some light has a wavelength, relative to M31, of 480 nm. What is its wavelength, relative to us?
4. A quasar emits electromagnetic radiation at a wavelength of 121.6 nm. If, relative to us, this wavelength is red-shifted 0.2 nm, what is the velocity of recession of the quasar?
Worked Solutions
# A-level Physics (Advancing Physics)/Electric Field/Worked Solutions

1. Two metal plates are connected to a 9V battery with negligible internal resistance. If the plates are 10 cm apart, what is the electric field at either of the plates?
$E_{electric}={\frac {V_{electric}}{d}}={\frac {9}{0.1}}=90{\mbox{ Vm}}^{-1}$
2. What is the electric field at the midpoint between the plates?
The whole point of a uniform electric field is that the field does not change anywhere between the plates - at the midpoint, as anywhere, it is 90 NC$^{−1}$.
3. The charge on an electron is -1.6 x 10$^{−19}$ C. What is the electric field 1μm from a hydrogen nucleus?
$E_{electric}={\frac {Q}{4\pi \epsilon _{0}r^{2}}}={\frac {1.6\times 10^{-19}}{4\pi \times 8.85\times 10^{-12}\times (10^{-6})^{2}}}=1440{\mbox{ Vm}}^{-1}$
4. What is the direction of this field?
The hydrogen nucleus has a positive charge, so the electric field goes away from the nucleus (by convention).
5. A 2C charge is placed 1m from a -1C charge. At what point will the electric field be 0?
Define the distance r as shown:
${\frac {2}{4\pi \epsilon _{0}(1+r)^{2}}}={\frac {1}{4\pi \epsilon _{0}r^{2}}}$
${\frac {2}{(1+r)^{2}}}={\frac {1}{r^{2}}}$
${\frac {(1+r)^{2}}{2}}=r^{2}$
$(1+r)^{2}=2r^{2}$
$1+r=r{\sqrt {2}}$
$r({\sqrt {2}}-1)=1$
$r={\frac {1}{{\sqrt {2}}-1}}={\frac {{\sqrt {2}}+1}{({\sqrt {2}}-1)({\sqrt {2}}+1)}}={\sqrt {2}}+1\approx 2.41{\mbox{ m}}$
# A-level Physics (Advancing Physics)/Electric Force/Worked Solutions

e = 1.6 x 10$^{−19}$C
1. A positron (charge +e) is 1 μm from a lithium nucleus (charge +3e). What is the magnitude of the force acting on each of the particles? In what direction is it acting?
$F_{electric}={\frac {kQq}{r^{2}}}={\frac {8.99\times 10^{9}\times 1\times 3\times (1.6\times 10^{-19})^{2}}{(10^{-6})^{2}}}=6.90\times 10^{-16}{\mbox{ N}}$
This may seem tiny, but the mass of an electron is also tiny, so the acceleration is considerable.
2. An electron is 1mm from the positively charged plate in a uniform electric field. The potential difference between the plates is 20V, and the plates are 10 cm apart. What force is acting on the electron? In what direction?
The 1mm is a red herring.
$F_{electric}={\frac {qV}{d}}={\frac {1.6\times 10^{-19}\times 20}{0.1}}=3.2\times 10^{-17}{\mbox{ N}}$
3. The acceleration due to gravity around a point mass is constant, irrespective of the mass of the objects it is acting on. The acceleration due to electricity around a point charge is not. Use Newton's Second Law (F=ma) to explain this.
$a={\frac {F}{m}}$
$a_{grav}={\frac {GMm}{mr^{2}}}={\frac {GM}{r^{2}}}$
$a_{electric}={\frac {kQq}{mr^{2}}}$
The m cancels for gravitational acceleration, but not for electrical acceleration, since charge does not provide a resistance to a force.
4. An insulator contains charged particles, even though the overall charge on the insulator is 0. Why is the insulator attracted by a nearby charge?
The electrons in the atoms can move within certain bounds. Being in an electric field means that they spend more time closer to a positive charge, and so the parts of the insulator which are attracted to the charge end up being closer to the charge than those which are repelled. This means that the attractive force on the insulator is greater than the repulsive force. This causes the insulator as a whole to be attracted to the charge if it is positive. If it is negative, the reverse happens, and the insulator is repelled.
5. Where in the charged conducting plates which create a uniform electric field would you expect to find the charge located? Why?
The charges are free to move within a conductor. The opposite charges in each plate are attracted to each other and try to move as close to each other as possible. So, they end up on the inside edge of the plates.
# A-level Physics (Advancing Physics)/Electric Potential
## Relationship to Electric Potential Energy

You will probably remember from AS (or even GCSE) that the energy U which flows along a wire is given by:
$U=Vq$,
where V is the potential difference between either end of the wire, and q is the amount of charge which flows. A simple rearrangement shows that:
$V={\frac {U}{q}}$
This potential difference is the same thing as electric potential. In a wire, the electric field is very simple. There are other electric fields, and in these fields as well, the electric potential is the electric potential energy per. unit charge. Electric potential energy between two point charges Q and q is given by:
$U={\frac {Qq}{4\pi \epsilon _{0}r}}$
So, the electric potential at a distance r from any point charge Q (ignoring other charges) is:
$V={\frac {Q}{4\pi \epsilon _{0}r}}$

## Relationship to Electric Field Strength

Electric potential is also the integral of electric field strength. This is why it is often called potential difference - it is an integral between two limits (two points in space) with respect to distance. So, the potential difference between two points a and b is:
$V_{ab}=\int _{a}^{b}E\;dx=\int _{a}^{b}{\frac {Q}{4\pi \epsilon _{0}x^{2}}}\;dx=\left[-{\frac {Q}{4\pi \epsilon _{0}x}}\right]_{a}^{b}$
But, if we define b as infinity and a as r:
$V=\left[-{\frac {Q}{4\pi \epsilon _{0}x}}\right]_{r}^{\infty }=-{\frac {Q}{4\pi \epsilon _{0}\infty }}-\left(-{\frac {Q}{4\pi \epsilon _{0}r}}\right)={\frac {Q}{4\pi \epsilon _{0}r}}$
So, the area under a graph of electric field strength against distance, between two points, is the potential difference between those two points.
For a uniform electric field, E is constant, so:
$V=\int _{r}^{\infty }E\;dx=Er$
In other words, V is proportional to r. If we double the distance between us and a point, the potential difference between us and that point will also double in a uniform electric field.

## Equipotentials

Equipotentials are a bit like contours on a map. Contours are lines which join up all the points which have the same height. Equipotentials join up all the points which have the same electric potential. They always run perpendicular to electric field lines. As the field lines get closer together, the equipotentials get closer together.

## Questions

ε$_0$ = 8.85 x 10$^{−12}$ Fm$^{−1}$
1. Draw a diagram of a uniform electric field between two plates, showing the field lines and the equipotentials.
2. Do the same for the electric field around a point charge.
3. The potential difference between two plates is 100V. What is the potential difference between a point halfway between the plates and one of the plates?
4. What is the electric potential at a point 0.2m from an alpha particle (charge on an electron = -1.6 x 10$^{−19}$C)?
5. What is the electric potential energy of an electron at the negative electrode of an electron gun if the potential difference between the electrodes is 10V?
Worked Solutions
# A-level Physics (Advancing Physics)/Electric Potential/Worked Solutions

ε$_0$ = 8.85 x 10$^{−12}$ Fm$^{−1}$
1. Draw a diagram of a uniform electric field between two plates, showing the field lines and the equipotentials.
2. Do the same for the electric field around a point charge.
3. The potential difference between two plates is 100V. What is the potential difference between a point halfway between the plates and one of the plates?
Half of 100 is 50V.
4. What is the electric potential at a point 0.2m from an alpha particle (charge on an electron = -1.6 x 10$^{−19}$C)?
The charge on an alpha particle is 3.2 x 10$^{−19}$C.
$V={\frac {Q}{4\pi \epsilon _{0}r}}={\frac {3.2\times 10^{-19}}{4\pi \times 8.85\times 10^{-12}\times 0.2}}=1.44\times 10^{-8}{\mbox{ V}}$
5. What is the electric potential energy of an electron at the negative electrode of an electron gun if the potential difference between the electrodes is 10V?
$U=Vq=10\times -1.6\times 10^{-19}=-1.6\times 10^{-18}{\mbox{ J}}=-10{\mbox{ eV}}$
# A-level Physics (Advancing Physics)/Electric Potential Energy

Just as an object at a distance r from a sphere has gravitational potential energy, a charge at a distance r from another charge has electrical potential energy ε$_{elec}$. This is given by the formula:
$\epsilon _{elec}=V_{elec}q$,
where V$_{elec}$ is the potential difference between the two charges Q and q. In a uniform field, voltage is given by:
$V_{elec}=E_{elec}d$,
where d is distance, and E$_{elec}$ is electric field strength. Combining these two formulae, we get:
$\epsilon _{elec}=qE_{elec}d$
For the field around a point charge, the situation is different. By the same method, we get:
$\epsilon _{elec}={\frac {-kQq}{r}}$
If a charge loses electric potential energy, it must gain some other sort of energy. You should also note that force is the rate of change of energy with respect to distance, and that, therefore:
$\epsilon _{elec}=\int {F\;dr}$

## The Electronvolt

The electronvolt (eV) is a unit of energy equal to the charge of a proton or a positron. Its definition is the kinetic energy gained by an electron which has been accelerated through a potential difference of 1V:
1 eV = 1.6 x 10$^{−19}$ J
For example: If a proton has an energy of 5MeV then in Joules it will be = 5 x 10$^6$ x 1.6 x 10$^{−19}$ = 8 x 10$^{−13}$ J.
Using eV is an advantage when high energy particles are involved as in case of particle accelerators.

## Summary of Electric Fields

You should now know (if you did the electric fields section in the right order) about four attributes of electric fields: force, field strength, potential energy and potential. These can be summarised by the following table:
| Force $F_{elec}={\frac {-kQq}{r^{2}}}$ | → integrate → with respect to r | Potential Energy $\epsilon _{elec}={\frac {-kQq}{r}}$ |
| ↓ per. unit charge ↓ |||
| Field Strength $E_{elec}={\frac {-kQ}{r^{2}}}$ | → integrate → with respect to r | Potential $V_{elec}={\frac {-kQ}{r}}$ |
This table is very similar to that for gravitational fields. The only difference is that field strength and potential are per. unit charge, instead of per. unit mass. This means that field strength is not the same as acceleration. Remember that integrate means 'find the area under the graph' and differentiate (the reverse process) means 'find the gradient of the graph'.

## Questions

k = 8.99 x 10$^9$ Nm$^2$C$^{−2}$
1. Convert 5 x 10$^{−13}$ J to MeV.
2. Convert 0.9 GeV to J.
3. What is the potential energy of an electron at the negatively charged plate of a uniform electric field when the potential difference between the two plates is 100V?
4. What is the potential energy of a 2C charge 2 cm from a 0.5C charge?
5. What is represented by the gradient of a graph of electric potential energy against distance from some charge?
Worked Solutions
# A-level Physics (Advancing Physics)/Doppler Effect/Worked Solutions

1. M31 (the Andromeda galaxy) is approaching us at about 120kms$^{−1}$. What is its red-shift?
$z={\frac {v_{s}}{c}}={\frac {-120000}{300000000}}=-0.4\times 10^{-3}$
The minus sign is important! Andromeda is blue-shifted!
2. Some light from M31 reaches us with a wavelength of 590 nm. What is its wavelength, relative to M31?
$-0.0004={\frac {\Delta \lambda }{\lambda _{0}}}={\frac {\lambda -\lambda _{0}}{\lambda _{0}}}={\frac {\lambda }{\lambda _{0}}}-1={\frac {590\times 10^{-9}}{\lambda _{0}}}-1$
$0.9996={\frac {590\times 10^{-9}}{\lambda _{0}}}$
$\lambda _{0}={\frac {590\times 10^{-9}}{0.9996}}=590.23{\mbox{ nm}}$
3. Some light has a wavelength, relative to M31, of 480 nm. What is its wavelength, relative to us?
$0.9996={\frac {\lambda }{\lambda _{0}}}={\frac {\lambda }{480\times 10^{-9}}}$
$\lambda =0.9996\times 480\times 10^{-9}=479.808{\mbox{ nm}}$
4. A quasar emits electromagnetic radiation at a wavelength of 121.6 nm. If, relative to us, this wavelength is red-shifted 0.2 nm, what is the velocity of recession of the quasar?
${\frac {v_{s}}{c}}={\frac {\Delta \lambda }{\lambda _{0}}}$
${\frac {v_{s}}{3\times 10^{8}}}={\frac {0.2}{121.6}}=0.00164$
$v_{s}=3\times 10^{8}\times 0.00164=493{\mbox{ kms}}^{-1}$
However, this is about as high a velocity as we can use the classical Doppler effect for.
# A-level Physics (Advancing Physics)/Electric Force

Electric fields are caused by charge. This charge can be either positive or negative. Like charges repel each other, and opposite charges attract each other. If we have two point charges of charge Q and q respectively, and they are a distance r apart, the force on each of them is:
$F_{electric}={\frac {kQq}{r^{2}}}={\frac {Qq}{4\pi \epsilon _{0}r^{2}}}$,
where k and ε$_0$ are constants (k = 8.99 x 10$^9$ Nm$^2$C$^{−2}$,ε$_0$ = 8.85 x 10$^{−12}$C$^2$N$^{−1}$m$^{−2}$). This means that, twice as far away from the point charge, the force on another charge decreases by a factor of 4. Electric force around a point charge is very similar to gravitational force around a point mass.
A uniform electric field consists of two conducting plates. These plates are oppositely charged, and infinitely wide. Obviously, infinitely wide conducting plates do not exist, so uniform electric fields do not exist. However, fields which approximate uniform electric fields do exist, provided we look towards the middle of the plates, and the plates are not too far apart - at the ends, the formulae for uniform fields no longer apply.
The force on a charge in a uniform electric field is given by:
$F_{electric}={\frac {qV}{d}}$,
where V is the potential difference between the two plates, q is the charge of the point charge upon which the force is acting, and d is the distance between the two plates. This force remains constant as the charge travels within the electric field.

## Questions

e = 1.6 x 10$^{−19}$C
1. A positron (charge +e) is 1 μm from a lithium nucleus (charge +3e). What is the magnitude of the force acting on each of the particles? In what direction is it acting?
2. An electron is 1mm from the positively charged plate in a uniform electric field. The potential difference between the plates is 20V, and the plates are 10 cm apart. What force is acting on the electron? In what direction?
3. The acceleration due to gravity around a point mass is constant, irrespective of the mass of the objects it is acting on. The acceleration due to electricity around a point charge is not. Use Newton's Second Law (F=ma) to explain this.
4. An insulator contains charged particles, even though the overall charge on the insulator is 0. Why is the insulator attracted by a nearby charge?
5. Where in the charged conducting plates which create a uniform electric field would you expect to find the charge located? Why?
Worked Solutions
# Adventist Youth Honors Answer Book/Arts and Crafts/Stamps (Template)

Stamps (Template) :
| Arts and Crafts General Conference | Skill Level 2 |
| Arts and Crafts General Conference | Year of Introduction: 1933 |
# Adventist Youth Honors Answer Book/Arts and Crafts/Stamps

Stamps :
| Arts and Crafts General Conference See also Stamps - Advanced | Skill Level 2 |
| Arts and Crafts General Conference See also Stamps - Advanced | Year of Introduction: 1933 |
The answers to the Stamps honor are dependent on the country in which you live. To keep from cluttering up the answers with all these variants, we instead provide separate answer chapters for each country.
Stamps (Australia) Stamps (Canada) Stamps (European Union) Stamps (India) Stamps (Indonesia) Stamps (Philippines) Stamps (United Kingdom) Stamps (United States)
If your country is not listed here, please add it. When you create the new page, start by inserting this text:
This will insert the requirements and all the invariant answers that have been added to date. You can then start adding the rest of the answers and editing answers we thought were invariant.
You can edit the main template here, but keep in mind that it should only have generally applicable answers (i.e., country-invariant answers).
Honors with Region-dependent Answers
# Adventist Youth Honors Answer Book/Arts and Crafts/Stamps (Australia)

Stamps (Australia) :
| Arts and Crafts South Pacific Division | Skill Level 2 |
| Arts and Crafts South Pacific Division | Year of Introduction: 1933 |
# Adventist Youth Honors Answer Book/Arts and Crafts/Textile Painting

Textile Painting :
| Arts and Crafts General Conference | Skill Level 2 |
| Arts and Crafts General Conference | Year of Introduction: 1956 |
# Adventist Youth Honors Answer Book/Arts and Crafts/Soap Craft (South Pacific Division)

Soap Craft (South Pacific Division) :
| Arts and Crafts South Pacific Division | Skill Level 1 |
| Arts and Crafts South Pacific Division | Year of Introduction: 1964 |

## 1. What is soap and how is it made?

Soap is a natural fatty substance which has been reacted /mixed with lye.

## 4a What is a detergent?

A detergent is any type of cleaner which doesn't contain / have soap as its main ingredient.
# AQA Business Studies/Elasticity of Demand

Elasticity of Demand is the term used to describe the difference between the price of a product and the demand. Generally the higher the price of the product the more the demand will fall and vice versa, although there are exceptions to this where a high price is used to make a product appear more "exclusive". Businesses use the elasticity of demand to determine what price to set in order to maximise profitability. Some businesses operate within markets where the products are considered inelastic. For example, petrol is generally considered to be inelastic as if the price rises the demand is not affected greatly. This is because vehicle owners need petrol so the increase in sales would be greater than the loss in demand. Over the long term it is elastic as vehicle buyers will be encouraged to buy vehicles using alternative fuels, find different means of transport (e.g., train, bus) or better fuel consumption. Individual petrol stations are elastic as consumers will go where the prices are cheapest. Commodity products are also generally inelastic (e.g., bread). An example of elastic demand might be McDonalds in the fast food market. They are in a highly competitive market and raising their prices too high would lead to a loss in demand greater than the increase in sales. Businesses will raise prices as high as possible if it demand is inelastic until it is so high it becomes elastic, whereas elastic products should be lowered to gain better profits caused by increase in demand.