Electric fields (DONE)

Question 1

What is an electric field?

Answer 1

• An electric field is a region where charged particles experience a force.

Question 2

What are common examples of electric fields?

Answer 2

• a polythene rod or a duster picking up small bits of paper.
• holding a charged rod close to a stream of water causing the water to bend towards the rod.

Question 3

What is the electric field strength?

Answer 3

• The electric field strength is the force experienced per unit positive charge.
• therefore E = F/Q

Question 4

Why is the electric field strength force per unit POSITIVE charge?

Answer 4

• This is because unlike gravity which is only ever attractive an electric field can be repelling or attracting.

Question 5

Electric field strength units?

Answer 5

• newtons per coulomb NC^-1

Question 6

Is the electric field strength a scalar or vector quantity?

Answer 6

• the electric field strength has a size and also a direction as it is the direction which a positive charge moves.
• this means the electric field strength is a vector quantity.

Question 7

How would you tell the direction of electric field lines?

Answer 7

• if you placed a proton in an electric field, whichever way the positively charged proton moves is the direction of the electric field.

Question 8

What experiment can be used to investigate electric field lines?

Answer 8

• you can use an EHT which is a high voltage source which goes up to around 5000V.
• you need to use shrouded leads to prevent risk of electrocution.
• place some caster oil which is an insulator, in a dish making a surface between the 2 plates which are placed on top of the liquid.
• One of the plates is positive and one is negative.
• In order to investigate the electric field, some semolina is placed in the caster oil.
• When you apply a really strong electric field one plate becomes positively charged and the other becomes negative.
• Because the semolina sits on top of the oil the grains line up and show the direction of the electric field.
• When all of the semolina is between the 2 positive and negative plates it lines up showing the field lines go from one positive plate to the negative plate.
• You can also do this with a point charge, the lines will radiate from the point charge to the opposite plate.
• Finally you can use a point charge with an opposite charge around the outside, this shows the semolina lining up in straight lines.

Question 9

How can you draw electric field lines accurately?

Answer 9

• around a positive point charge we draw 8 straight field lines acting away from the positive point charge.
• All of the lines should meet the point charge at 90 degrees.
• When the field lines are closer together this is where we have a stronger field.
• positive and negative field lines should attract and connect eventually except from the directly opposite central line however this will not be able to be shown in a small diagram.
• If a positive charge is placed in the field it will flow towards the negative charge and away from the positive.
• If you have 2 positive point charges the field lines will never cross, and if a positive charge is placed in the field it will be repelled from both.
• If you have a positive point charge and a negatively charged plate the field lines will all reach the plate at 90 degrees.
• And if you had a positive and a negative plate you will have a uniform field in the middle.

Question 10

What can we use coulombs law to calculate in electric fields?

Answer 10

• If we had 2 positive protons near one another we could calculate the force between them.
• moving them closer together will increase the repulsive force.

Question 11

How do you derive the equation for the force between 2 charged particles using coulombs law?

Answer 11

• The force between the 2 objects will be dependant on the size of charge Q and q therefore Force is proportional to charge Q and q.
• The force also depends on how close together the charges are, the separation distance from the centre of Q to q is represented by the symbol r.
• The force is proportional to 1/(r^2) as it is not just in one dimension we have to show the field strength lines radiating in all directions.
• If we combine the 3 proportional equations what we can say is that the force is proportional to (Qq)/(r^2)
• we now need a constant of proportionality which = 1/4pieE0
• Therefore F = [1/4pieE0] x [(Qq/(r^2)]
• if you have a positive charge Q and another positive charge q, the force between them is a positive force meaning it is repelling.
• If one of the charges is negative that means the force between the 2 objects is negative meaning it is attractive.
• Epsilon 0 which is represented by the symbol E0 is the permittivity of free space with a value of 8.85 x 10^-12 f

Question 12

What are radial electric fields?

Answer 12

• Radial fields are the fields around a point object.
• They are very similar to a gravitational field around the earth.
• The electric field strength, E = F/Q
• The force between 2 objects, F = [1/4pieE0] x [Qq/r^2]

Question 13

How can you calculate the electric field strength in a radial field at the point q?

Answer 13

• The electric field strength is going to be equal to the force (between the 2 charges F = [1/4pieE0] x [Qq/r^2]) per unit charge, q.
• This therefore cancels down to E = [1/4piwE0] x Q/r^2
• This means if we had double the charge we would have double the force on the object however the force per unit charge will be the same size therefore it doesn’t matter about the size of the charge.
• This tells us that the size of the electric field E is proportional to charge Q.
• So if you have a bigger charge causing an electric field, the field will be larger.
• The electric field strength, E is proportional to 1/r^3

Question 14

What will happen to a unit charge in a uniform electric field?

Answer 14

• if you have 2 metal plates and the bottom plate has a positive potential difference and the top plate has 0 potential difference, between them there will be an electric field.
• if you place a charged particle between the 2 plates it will experience a force.
• the reason for this is because between the 2 plates we have a uniform electric field.
• for example if you had an electron on the positive bottom plate, this is where the electron wants to be but what we can do is move it around.
• In order to move the electron away from the positive plate it is attracted to you need to do some work on it.

Question 15

How can you derive an equation for uniform electric field strength?

Answer 15

• Perhaps the test charge in the field has a charge of q.
• If you look at the equation for the volt, V = W/q, you can rearrange this to work done, W = Vq.
• this means it is going to take a greater amount of work to move the charge if it has a bigger charge or if the plates have a bigger potential difference.
• electric field strength, E = F/q
• meaning that the force, F = Eq.
• this means if you have a bigger electric field strength or charge, the force pulling the charge to the positive plate is going to be larger.
• You can then substitute F = Eq and W = Vq into the equation for work done W = Fd
• This results in Vq = Eqd
• This can be cancelled to get V = Ed
• Or E = V/d which is the electric field strength = potential difference/distance moved in direction of force.
• this means if the separation distance between plates is larger the size of the electric field strength is smaller.

Question 16

What is a parallel plate capacitor and how does capacitance change?

Answer 16

• a capacitor is a device which stores charge.
• it consists of 2 oppositely charged vertical plates separated by an insulator which is often a dielectric.
• These 2 plates have a uniform field between them.
• the capacitance, C is proportional to the area of the plates.
• the capacitance, C is proportional to 1/d where d is the separation of the plates.
• We can therefore day that the capacitance is proportional to the A/d which is the area of the plates/the separation between them.

Question 17

What is the equation for capacitance and how do you calculate the permittivity of the insulator?

Answer 17

• capacitance is proportional to A/d which is the area of the plates/the separation between them.
• We have to put a constant of proportionality into the equation therefore C = (EA)/d.
• E (epsilon) = permittivity of the insulator
• E = ErEo
• where Eo = the permittivity of free space = 8.85 x 10^-12 Fm^-1 (It is how capable a vacuum is of emitting electric field line).
• and Er = the relative permittivity (which depends on the material between the plates).

Question 18

How can we investigate the motion of charged particles in electric fields?

Answer 18

• Perhaps we have an electron held between 1 neutral plate and another positively charged plate.
• The electron will accelerate towards the positive plate and the motion has a lot of similarities to how an object moves in a gravitational field.
• The electron has a charge q and a mass m, and the electric field which the electron is held in has a strength E.
• You can use SUVAT to investigate this motion of the electron.
• you know the intial velocity, u = 0
• The acceleration will be equal to the force/mass, therefore we need to find the force of the electric field which can be found using different ways.
• You could use F = Eq or E = V/d to find the acceleration.
• Using the acceleration you can then work out the final velocity.
• We could also fire electrons into the electric field and investigate the motion.
• you can split the motion into horizontal and vertical components.
• The uniform force of the electric field is constant and therefore the acceleration of the electron towards the positive plate is constant.
• So as it moves through the field it has the same horizontal velocity but its vertical velocity will increase.

Question 19

How can you calculate the capacitance of an isolated sphere?

Answer 19

• A capacitor doesn’t have to be 2 plates separated by a gap, we can look at the capacitance around an object which is charged.
• Perhaps what we have is the top of a van der graaf generator, we can look at the capacitance of a charged sphere.
• it is important that we consider just the sphere and nothing else around it so we have an isolated sphere.
• The sphere has a total charge Q, and it also has a radius of R.
• We can work out the electric potential at its surface, V = Q/(4pieE0R)
• We can rearrange this to Q = V4pieE0R
• Capacitance, C = Q/V
• Therefore we can substitute Q into C = Q/V
• Through cancelling we then get C = 4pieE0R
• This means that the capacitance is proportional to the size of the sphere.

Question 20

What is electric potential energy?

Answer 20

• Electric potential energy is the work done to bring the particles from infinity to a separation of r.

Question 21

How is electric potential energy created?

Answer 21

• If you have 2 charged particles of the same charge for example 2 protons, you can move them from a distance infinity where there is no force between them and you can start to move the protons closer together.
• The more you push them the more they repel so the more force you need to apply to get them closer together.
• This is similar to how you compress a spring and it stores elastic potential energy.

Question 22

What relationship does the graph of force against separation show (electric potential energy)?

Answer 22

• Because the force, f is repulsive it means it must be a positive value throughout.
• As the distance, r gets smaller the force, f will get larger.
• The line tends towards 0 as r reaches infinity.
• We can look at the work done required to push the protons together.
• because we have a force distance graph the work done is equal to the area under the graph.
• Therefore if you brought the protons to a distance r away from each other, the work done in bringing it from infinity to that point is the area to the right of the r.

Question 23

What is the equation for electric potential energy?

Answer 23

• E = (Qq)/(4pieE0r)
• If you have 2 particles of the same charge this is going to be a positive energy whereas if the charges are attractive the energy is negative as we need to do work to separate the particles.

Question 24

What is electric potential?

Answer 24

• The electric potential is the work done per unit charge to bring a positive charge to that point from infinity.

Question 25

What is the equation for electric potential?

Answer 25

• We know that the electric potential energy E = (Qq)/(4pieE0r)
• However the electric potential looks at the work done per unit charge.
• electric potential, V = E/q
• We can replace the E term with (Qq)/(4pieE0r) making the equation V = [(Qq)/(4pieE0r)]/q
• Therefore the q on the top and the bottom cancel each other out.
• The final equation V = (Q)/(4pieE0r)

Question 26

What are the units for electric potential?

Answer 26

• This can have the units JC^-1 or V, the volt is the units for the potential difference which is the work done per unit charge.