7.1 - Electric Fields

Question 1

How do we push charges - give an example

Answer 1

Many machines function through the use of fast moving charged particles. For example, in a hospital x ray machine high speed electrons are crashed into a metal target in order to produce the x rays. So how do we cause the electrons to move at high speed? A region of spaced that will cause charged particles to accelerate is said to have an electric field

Question 2

How can we visualise the forces caused by an electric field

Answer 2

We draw electric field lines which show the direction a positively charged particle will be pushed by the force the field produces

Question 3

What does the spacing of field lines mean

Answer 3

As with magnetic fields patterns - indeed as with all field patterns - the close together the lines are, the stronger the field is

A stronger field causes stronger forces

Question 4

A stronger field causes…

Answer 4

…..Stronger forces

Question 5

How can we calculate the force F, that a charged particle will feel

Answer 5

It’s the electric field strength (E) multiplied by the amount of charge in coulombs (Q) as given by the equation

F = EQ

Where F is the force, Newton’s
E is electric field strength, in Vm^-1 or NC^-1
Q is charge on the particle

Question 6

From this force equation, F= EQ, how can we calculate how quickly a charge would accelerate

Answer 6

Newton’s second law is that F = ma, we can combine these equations

F = EQ = ma
a = EQ/m

Question 7

What’s the electric potential

Answer 7

Every location within a field gives a charged particle a certain electric potential energy per unit charge. This is called electric potential

Question 8

What is the equation for electric potential

Answer 8

V = E(subscript p)/ Q

This is similar to the relationship between force and field strength. Field strength is the force felt by a charged particle per unit of charge, expressed by the equation E = F/Q

The difference between that and a new location that the electron might move to is called the potential difference through which the electron moves.

Question 9

What is potential difference

Answer 9

The difference between that and a new location that the electron might move to is called the potential difference through which the electron moves.

We previously defined potential difference for a device in an electrical circuit as the energy transferred per coulomb of charge passing through the device. In an electric field, we can follow exactly the same idea in order to find out how much kinetic energy a charged particle will gain by moving within the field. This is given by the equation:
E(subscript k) = VQ

Question 10

How can we calculate the kinetic energy of a charged particle

Answer 10

The equation

E(subscript k) = V x Q

E(subscript k) = 1/2 x m x v^2

Question 11

Where can an electric field exists

Answer 11

An electric field exists between any objects which are at different electrical potentials. Thus, if we connect parallel metal plates to a power supply, we can set up a potential difference, and therefore an electric field, between them.

Question 12

The field is uniform if…

Answer 12

It’s field lines are parallel across the whole field

Question 13

What is the strength of a uniform electric field

Answer 13

It’s a measure of how rapidly it changes the potential over distance. The equation which describes this divides the potential difference, V, by the distance over which the potential difference exists

E = V/d

Question 14

What two ways can a uniform electric field strength be defined as

Answer 14

E = F/Q

E = V/d

This gives rise to two equally valid SI units for electric field strength, NC^-1 and Vm^-1

Question 15

How can we investigate electric fields

Answer 15

Make sure you know how to use an EHT power supply safely before conducting this experiment. Only use an EHT power supply provided by the technicians.

You can investigate the shapes of electric fields using an EHT power supply to provide a potential difference, and castor oil with floating semolina to show the field lines. The semolina becomes slightly charged, and the charges on the semolina grains cause them to line up. Thereby showing the lines of action of the forces produced by the field. Try it with different shapes electrodes to see uniform and non-uniform fields.

Question 16

What are equipotentials

Answer 16

As we move through an electric field, the extrications potential changes from place to place. Those locations that all have the same potential can be connected by lines called equipotentials

These are very much like the contours on a map, which are lines of equal height, and thus indicate lines of gravitational equipotential

The field will always be perpendicular to the equipotential lines, as a field is defined as a region which changes the potential.

How close the equipotentials are indicates the strength of the electric field, as this shows how rapidly the electric potential is changing.

Question 17

What can field lines and equipotentials not do

Answer 17

CROSS they can’t CROSS
NEVER EVER NEVER

But never say never;)

YES DO SAY NEVER

Question 18

How can we investigate equipotentials

Answer 18

You can investigate how electrical potential varies across an electric field by measuring the voltage between zero and the point within the field that you are interested in. A simple experiment using conducting paper and a low voltage power supply to set up an electric field will allow you to map out equipotentials, and this will enable you to produce a picture of the electric field.

You can try it with differently shaped electrodes in order to see how they change the field shape.

Question 19

Define an electric field

Answer 19

An electric field is a region of space that will cause charged particles to experience a force

Miss Says “small positive charged particle” instead of charged particles

Question 20

Define electric field lines

Answer 20

Electric field lines are imaginary visualisations of the region of influence of an electric field.

Question 21

Define electric field strength

Answer 21

Is the force to charge ratio for a charged particle experiencing a force due to an electric field.

Question 22

Define potential

Answer 22

Potential is a measure of possible field energy in a given location within that field. Electric potential is the energy per unit charge at that location.

Question 23

Define potential difference

Answer 23

Potential difference is the change in potential between two locations in a given field.

Question 24

Define equipotentials

Answer 24

Equipotentials are positions within a field with zero potential difference between them

Question 25

What can we use to find the ratio of charge/mass for particles

Answer 25

Scientists can use a mass spectrometer

Question 26

What can the ratio of charge/mass show

Answer 26

This is a fundamental property of a charged particle and identifies it uniquely, but until the determination of the value of the charge on an electron, it was impossible to break down the ratio and find the mass of that particle.

In 1909, Robert Millikan developed an experiment which determined the charge on a single electron

Question 27

Simply, what did Robert Millikan do

Answer 27

Developed an experiment which determined the charge on a single electron

Question 28

What was the experiment that Millikan did to find the charge on a single electron

Answer 28

A charged droplet of oil has its weight balanced by the force from a uniform electric field. When oil is squirted into the upper chamber from the vaporiser, friction gives the droplets an electrostatic charge. This will be some unknown multiple of the charge on an electron, because electrons have been added or removed due to friction. As the drops fall under gravity, some will go through the anode and enter the uniform field created between the charged plates. If the field is switched off, they will continue to fall at their terminal velocity. Check diagram on page 37

Question 29

How did Millikan calculate the charge on an electron with data from the experiment

Answer 29

For Millikans oil drops, the density of air in the chamber is so low that up thrust is generally insignificant. (Although it would have been considered if we wanted to do really accurate calculations)Under terminal velocity, the weight equals the viscous drag force (using stokes law)

mg = 6 x pi x n x v(subscript term) x r
Where n is the viscosity of air, and r is the radius of the drop.

When held stationary by switching on the electric field and adjusting the potential, V, until the drop stands still:
weight = electric force

mg = EQ = VQ/d

By eliminating the weight from the two situations, this means that

Q = 6 x pi x n x v(term) x r x d / V

Millikan could not measure r directly, so he had to eliminate it from the equations.
Further rearrangement of stokes law allows us to calculate Q.

He calculated the charge very accurately (within 1% accuracy)

Question 30

What is the electric field like around a point charge

Answer 30

It is a radial electric field, in the region around a positively charged sphere, or point charge such as a proton, the electric field will act outwards in all directions away from the centre of the sphere.

You will see that field lines get further apart as you move further away from the sphere

Question 31

What happens to field strength in racial fields

Answer 31

You will see field lines/arrows get further apart as you move further away from the sphere, indicating that the field strength reduces as you move away from the centre. This means that the distance between equipotentials also increases as you move further away from the centre.

The field is the means by which the potential changes, so if it is weaker, the potential changes less quickly

Question 32

What about the charged sphere in the centre of the field/ causing a field

Answer 32

Within a charged sphere, there is no electric field, as a charged particle would feel no resultant force on it the overall effect of all the charges on the sphere cancel out within the sphere itself.

Distance measurements for use in calculations should always be considered from the CENTRE OF a charged sphere

Question 33

Tell me about calculating radial electric field strength

Answer 33

We have seen that a radial field has its field lines spreading further apart simply due to geometry. This means that the equation for the field strength of a radial electric field must incorporate the weakening of the field with distance from the charge.

The expression for radial field strength at distance r from a charge, Q is:

E = Q/4 x pi x epsilon(subscript 0) x r^2

Strictly speaking this is only correct for a field that is produced in a vacuum lol fysics is sooo crazy😰❤️
As the value epsilon(subscript 0) - the permittivity if free space - is a constant which relates to the ability of the fabric of the universe to support electric fields. It’s value is epsilon(subscript 0) = 8.85 x 10^-12 F/m

Other substances, for example water may be better or worse at supporting electric fields, so an extra factor (the relative permittivity, epsilon(subscript r) would appear in the equation to account for this.

Air is considered to be near enough a vacuum that we use the equation with epsilon(subscript 0) for electric fields in air.

Question 34

Tell me about the permittivity of free space in the radial electric field strength equation.

Answer 34

Strictly speaking epsilon(subscript 0) is only correct for a field that is produced in a vacuum lol fysics is sooo crazy😰❤️
As the value epsilon(subscript 0) - the permittivity of free space - is a constant which relates to the ability of the fabric of the universe to support electric fields. It’s value is epsilon(subscript 0) = 8.85 x 10^-12 F/m

Other substances, for example water may be better or worse at supporting electric fields, so an extra factor (the relative permittivity, epsilon(subscript r) would appear in the equation to account for this.

Air is considered to be near enough a vacuum that we use the equation with epsilon(subscript 0) for electric fields in air.

Question 35

What does electric field strength tell us in radial fields

Answer 35

Electric field strength tells us how quickly the electric potential is changing. A stronger field will have the equipotentials closed together.

Question 36

Tell me about the potential in a radial electric field (equation)

Answer 36

A stronger field will have the equipotentials closer together. Thus equation states that the electric field strength, E, is equal to the rate of change of potential, V, with distance x

E = -dV/ dx

This leads to the expression for radial field potential at a distance r from a charge Q:

V = Q/4 x pi x epsilon(subscript 0) x r

You will see from the equation above that the position of 0 potential can be considered to be infinity

Question 37

Btw the constant in this equation:

E = Q/4 x pi x epsilon(subscript 0) x r^2

Answer 37

The constant expression 1/4 x pi x epsilon(subscript 0) is sometimes written as a single k (Coulombs constant). The value for k is:

k = 8.99 x 10^9 Nm^2C^-2

This means our expressions for radial field strength and potential can be written

E = kQ/r^2

V = kQ/r

Question 38

How can we combine electric fields

Answer 38

In a region where there are electric fields caused by more than one charged object, the overall field is the vector sum at each point of all the contributions from each field. If you imagine building the overall field up from the force effects of each contributing field, you can see that at every point you have to work out the resultant force, and in this way you know how a charged particle would be affected there. The sum of all of these individual force effects is the overall electric field.

Charge is particularly concentrated in regions around spikes or points on charged objects. This means that they have close field lines and the field will be strong around them.

For this reason, lightning conductors are spiked: the concentrated charge will more strongly attract opposite charges, so lightning is more likely to hit the lightning conductor than the building being protected.

In fact, the field around a spiked lightning conductor is often strong enough that it can cause charge leakage through the conductor before charge builds up to a point to produce a lightning strike. The probability of lightning occurring is reduced, further protecting the building from these dangers.

Question 39

Tell me about the fields in a charged particle interaction

Answer 39

The attraction between a proton and an electron can be imagine as the proton creating an electric field because of its particle charge, and the electron feeling a force produced by the protons field. It could also be thought of the other way around, with the proton feeling a force caused by the electrons electric field.

Question 40

What is Coulomb’s law

Answer 40

F = KQ1Q2/r^2

F = force between two charged particles
Q1 and Q2 are charged particles, separated by a distance r.

Note that for charged spheres, the calculations of the force between them is also given by Coulomb’s law, but the distance is measured between the centres of the spheres.

Take care to correctly include the charge signs in the calculation of force between charges, as the sign of the final answer will tell you wether they attract or repel. A negative force will be attractive, as the force is then in the opposite direction to that in which the separation was measured.

Question 41

How can we investigate Coulomb’s law

Method 1 (there are 2 ways)

Answer 41

A pair of insulated metal uses spheres can be charged using a Van de Graaff generator, or simply by induction using a charged plastic rod.
Clamping the spheres close to each other will cause a force between them which can be measured using an electronic balance. By adjusting the distance of separation and measuring the force at each distance, this apparatus can be used to confirm that Coulomb’s law follows an inverse square law.

Question 42

Tell me about everyday levitation - can we really touch anything?

Answer 42

All materials are made of atoms, and the outer part of atoms is considered to be the region of electron orbits - the location of the negative charge. We can therefore imagine the top of any surface to consist of a sheet of negative charges. So, the surface of the floor and bottom of a mans shoe will both be planes covered with negative charge. The repulsion between these two surface will cause the man to levitate ever so slightly above the floor, rather than actually touching it.

Question 43

How can we verify Coulomb’s law

Method 2

Answer 43

In 1785, Charles Augustine de Coulomb published a paper in which he provided experimental evidence for the equation to calculate the force between two charges. His experiments involved measuring the force between charges spheres on a torsion (rotating) balance and fixed charged spheres. This method can be simplified to carry out in a school lab.

By reducing the separation of the spheres, you will see the angle of the lighter hanging ball increase. You can investigate the force between charged spheres and confirm that there is an inverse square law involved, and that the force is proportional to the product of the charges.

Check diagram on page 43, by resolving and equating the horizontal and vertical components, we can develop a method for verifying Coulomb’s law.

For a fixed amount of charge on the spheres, you could vary r and measure d, from which a plot of 1/r^2 should produce a straight line. It is very difficult to specifically control the amount of charge on the spheres, but having charged both up, touching one with a third, uncharged, sphere can remove half the charge on that one. Doing this several times, and measuring d in each case, will allow you to confirm that F(subscript coulomb) is proportional to the product of the two charges.