Key Areas covered The relationship between potential difference, work and charge gives the definition of the volt. Calculation of the speed of a charged particle accelerated by an electric field.

What we will do today: State that in an electric field, an electric charge experiences a force. Carry out calculations on the above

Potential Difference

When a charge Q is moved in an electric field, work W is done. If one joule of work is done moving one coulomb of charge between two points in an electric field, the potential difference between the two points is one volt.

Potential Difference (p.d.) = Work Done Charge V = W W=QV Q What is meant by potential difference? The potential difference (voltage) V between two points in an electric field is a measure of the work done in moving one coulomb of charge between the two points. 1 volt = 1 joule per coulomb 1 V = 1 JC-1 On data sheet

Example + - - + - + - A +ve charge of 3 μC is moved as shown, between a p.d. of 10 V. (a) Calculate the potential energy gained. (b) If the charge is released, state the energy change. (c) How much kinetic energy is gained on reaching the negative plate? (a) W = QV = 3 x 10-6 x 10 = 3 x 10-5 J (b) Electric potential to kinetic energy (c) Conservation of energy: It will be the same i.e. 3 x 10-5 J + +Q + - + - + - + -

Change in Pot. Energy = Work Done Note: In this section W will be used for Work Done, i.e. energy transferred. From N5: Change in Pot. Energy = Work Done = gravitational force x height = mgh Note that there are similarities between electric and gravitational fields.

We can infer that in electrical terms: Change in electrical potential energy = electric force x distance If force is in direction of electric field, energy appears as kinetic energy. If force is against the direction of the electric field, energy is stored as potential energy.

In the previous example, when the +ve charge was released, EP EK. i.e. QV = ½mv²

Info from data sheet Many questions may require reference to the data sheet at the beginning of the exam paper.

Example An electron is accelerated (from rest) through a potential difference of 200 V. Calculate: (a) the kinetic energy gained (b) the final speed of the electron. [me = 9.11 x 10-31 kg, Qe = 1.6 x 10-19 C] (a) Ek = ½mv² = QV = 1.6 x 10-19 x 200 = 3.2 x 10-17 J (b) ½mv² = 3.2 x 10-17 v² = (3.2 x 10-17 x 2) / m = (3.2 x 10-17 x 2) / 9.11 x 10-31 v = 8.4 x 106 ms-1

Common PS questions Very often the question may change one of two possible factors and ask you to comment on the possible outcome. You should refer to the equation and see if it affected by the change – QV = ½ mv2 The distance of the field: This would not affect the velocity of the charge as it is only dependent on the variables in the equation – Q, V and m. These have not changed so neither does velocity. The charge of the particle: This would affect the velocity of the charge. A smaller Q means smaller velocity and a larger Q means larger velocity.

Past Paper Questions

2003 Qu: 8

2005 Qu: 7

2005 Qu: 8a

2008 Qu: 8

2009 Qu: 7a

2010 Qu: 7

2006 Qu: 24

2007

Revised Higher 2012 (1/2)

Revised Higher 2012 (2/2)

CfE Specimen Paper