5.1 Electric Fields

The diagram shows an air filter which uses charged collecting plates to remove dust from the air of a workshop.

The air intake passes through a charged, ionising grid which attracts dust particles, cleaning the air which is then returned back into the workshop.

A dust particle of mass 6.7 × 10^–15 kg enters the region between the collecting plates travelling horizontally with an initial velocity of 11 m s^–1. The particle carries a charge of 2.6 × 10^–18 C.

 Assume that the dust particles move horizontally between the plates.

Determine the electrostatic force acting on the particle.

Some particles are not caught by the air filter, but pass straight through. Others are caught by the filter. The particles are identical in mass and charge, and they all travel parallel to the plane of the plates. The plates are initially completely clean. Assume the particles are evenly vertically distributed.

Deduce the percentage of dust particles which will be 'trapped' by the negatively charged plate. Ignore the effect of gravity.  

As the air filter operates, there is a build up of particles on the negative plates. The gap between the plates therefore becomes narrower, by up to 10% of its initial height.

Discuss whether this narrowing makes the filter more or less effective at removing dust particles.

Two charged objects X and Y are made to circle a point O. X and Y are at a distance, d = 1.8 × 10⁻⁸ m and they have equal masses, where m = 1.7 × 10⁻⁹ kg. The objects carry an equal but opposite charge, where the magnitude q = 3.2 × 10⁻¹⁹ C.

For this motion calculate

The particles X and Y in part (a) are replaced with a gold nucleus , and an alpha particle.

The alpha particle and gold nucleus are at rest at a distance where the electric fields only just interact with each other.

For the axes shown sketch the graph of electric potential V against distance along the straight line between the charges.

An experiment to determine the charge on an electron is shown.   

Negatively charged oil drops are sprayed into a region above two parallel metal plates which are separated by a distance, d. The oil drops enter the region between the plates.

A potential difference V is applied which causes an electric field to be set up between the plates.

(i)

Using the sketch below, which shows one oil drop falling between the plates, show the electric field between the plates.

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(ii)

Hence or otherwise explain why the oil drop stops falling when V is increased.

[2]

The oil drop has mass = m and charge = q. The distance between the plates = 2.5 cm.

The oil drop stops falling when potential difference, V = 5000 V

Determine the charge to mass ratio of the oil drop.

Two oil drops are suspended between the plates at the same time. The oil drops can be considered as identical point charges with mass 1 × 10⁻¹³ kg which are spaced 2.2 mm apart.

Calculate the electrostatic force between the drops.

For the oil drops in part (c)

Describe and explain the expected observations as the potential difference increases above 5000 V, using a mathematical expression to justify your answer.

A uniform copper wire contains 5.0 × 10²³ electrons and has a current of 2.0 A flowing through it.

Calculate the time it will take all the electrons present in the wire at one instant to come out of the end.

The wire in part (a) is 5.0 m in length and has a diameter of 1.22 mm. 

Calculate the electron density in the copper wire.

Using the calculated values from parts (a) and (b)

An integrated circuit uses thin strips of gold and silicon as connectors and resistors respectively.

A strip of gold, of cross-sectional area 2.0 × 10^–6 m² has a charge carrier density of  7.0 × 10²⁸ m^–3 and a current of 8.5 mA.

Calculate the charge carrier drift speed for gold.

An approximate value for the charge carrier drift speed for a sample of silicon of the same dimensions, carrying the same current, would be 0.20 m s^–1.

Compare this value with the one you obtained in part (a) for gold and explain the reason for the difference between the two drift speeds.

In another integrated circuit a current of 2.0 A flows through a resistor for 90 minutes.

Determine the number of electrons that pass a point in the resistor this time.

The current in part (c) flows across a potential difference of 12 V.

Using your answer to part (c), calculate the total energy transferred in the integrated circuit.

A parallel-plate capacitor is an electrical component that stores electric charge.

It is set up by connecting two metal plates to a power supply.

Label: 

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State, for each of the scenarios below, whether the electric field strength between the metal plates increases, decreases, or stays constant:

A free electron finds itself incident in the space between the metal plates and is deflected as it moves between them.

The magnitude of the electric field strength is 200 N C^–1. Calculate the magnitude of the electron’s acceleration in the space between the plates.

Explain the shape of the path shown in part (c).

State Coulomb’s law in words.

In simple models of the hydrogen atom, an electron is in a circular orbit around the proton.

The magnitude of the force between the proton and the electron is 5.8 × 10^–9 N.

Calculate:

   (i)        the orbital radius of the electron.

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The gravitational field strength g due to the proton at any point in the electron’s orbit is given by the equation:

                                                                   g = G

where  is the proton mass, r is the orbital radius and G is the gravitational constant.

Show that the ratio of the gravitational field strength to the electric field strength due to the proton at any point in the electron’s orbit is of the order 10^–28.  

Ionisation is the process of removing an outer shell electron from an atom, so it is transferred from its orbit to a point where the potential is zero.

The potential difference between the electron’s orbit in a hydrogen atom and this point is about 3.4 V.

Calculate the gain in potential energy of an orbiting electron in a hydrogen atom if it is ionised.

When a copper wire is exposed to an electric field, a current is detected in it.

Explain, with reference to charge carriers, why there is a current detected in the wire.

Calculate the drift speed of the charge carriers in the copper wire if a potential difference of 4.0 V is applied to it.

The following data are available for the wire:

               density of free electrons = 8.5 × 10²⁸ m^–3

               resistance = 25 Ω

               diameter = 0.8 mm

A defect is discovered in the wire. This causes the cross-sectional area to increase by 42% at a point X along its length.         

Calculate the new drift speed of charge carriers in the copper wire at point X if the same potential difference is applied as that in part (b).

Explain how you arrived at your answer.

When the applied potential difference is removed, the current in the copper wire falls to zero. However, individual charge carriers move within the wire at thermal speeds which are often orders of magnitude above drift speeds.

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