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1. Uniform Electric Fields

2. Electric Fields Content • Concept of an electric field • Force between point charges • Electric field of a point charge • Uniform electric fields Learning Outcomes Candidates should be able to: (a) understand an electric field as an example of a field of force and define, recall and use electric field strength as force per unit positive charge. (b) use field lines to represent an electric field. (c) recall and use Coulomb’s law for point charges in a vacuum in the form F = kQ1Q2 / r2, where k =1 / 4πε0. (d) recall and use E = kQ / r 2 for the electric field strength of a point charge. (e) recall and use E = V/d for the magnitude of the uniform electric field strength between charged parallel plates. (f) recognise the similarities of, and differences between, electric fields and gravitational fields.

3. Electric Field Phenomena • Rubbed balloon sticking to wall • Lightning • Photocopying • Laser printing • Flue-ash precipitation • Spray Painting

4. Electric Field Strength • Gravitational fields are concerned with masses, whereas electric fields are concerned with charges. An electric field exists in a region of space in which a stationary charge experiences a force. • Electric charges create an electric field in the space around them. Unlike gravitational forces, which are always attractive, electric forces can be attractive or repulsive. • The electric field strengthE at a point is the force experienced per unit positive charge on a point charge placed at that point. E = F / Q • Where F is the electric force experienced by a particle of charge +Q. Like force, electric field strength is a vector. Rearranging this equation gives F = EQ • Electric field strength has units N C-1.

5. Problems Data k = 1 / 4πεo = 9.0 x 109 N m2 C-2 Charge of electron = 1.6 x 10-19 C • Where the field strength is 1000 NC-1, what is the force on a 1 C charge? On an electron? F = EQ = 1000 x 1 = 1000 N F = EQ = 1000 x 1.6 x 10-19 = 1.6 x 10-16 N • A charged sphere is placed in a field of strength 3 x 104 N C-1. If it experiences a force of 15 N, what is the charge on the sphere? Q = F/E = 15 /(3 x 104) = 5 x 10-4 C 3.What is the field strength if an electron experiences a force of 4.8 x 10-14 N? E = F/Q = 4.8 x 10-14 / (1.6 x 10-19) = 3 x 105 N C-1

6. Electric Field Patterns Electric field patterns are mapped out using electric field lines. They have the following properties. • They never start or stop in empty space only from a charge or at “infinity.” • They never touch or cross (otherwise a charge placed at that point would experience forces in different directions). • They have direction. The tangent to an electric field line gives the direction in which a positive point charge placed at that point would move. • The density of electric field lines indicates the strength of the electric field. Closely spaced electric field lines indicates greater electric field strength. • Parallel and equally spaced electric field line indicate an electric field of constant field strength (a uniform field).

7. Demonstration Semolina grains sprinkled onto the surface of castor oil align themselves in the direction of the electric field caused by differently shaped electrodes/conductors. • 0.5 cm castor oil + thin layer of semolina • EHT power supply, 3000-4000 V

8. Practical • Plotting equipotentials

9. Two parallel conductors.

10. - An isolated metal ring.

11. Neutral Point The force exerted on a charge here would be exactly equal and opposite

12. Worksheet • Field Lines and Equipotentials

13. F F +Q1 +Q2 r Coulomb’s Law • The magnitude of the force between the charged spheres is indicated by the deflection of each sphere from the vertical plane. • The force can be attractive (unlike charges) or repulsive (like charges). • The force increases as the separation decreases. • The force increases as the charge increases. • F α Q1Q2 / r2 • F = kQ1Q2 / r2 • Where k = 1 / 4πεo • Where εo is the permittivity (how well a material supports an electric field) of free space = 8.85 x 10-12 F m-1 • Coulomb’s Law of electrostatics states that the magnitude of the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of their separation. –ve F = attraction +ve F = repulsion

14. Worked Example • According to a simple model of the hydrogen atom, the mean separation between the proton and the electron is 5.0 x 10-11 m. Calculate the electric force acting on the electron. • e = -1.6 x 10-19 C • εo = 8.85 x 10-12 F m-1 • F = - 9.21 x 10-8 N • The minus sign denotes attraction between the proton and the electron.

15. What is the force of repulsion between two electrons held 1 m apart in a vacuum? By what factor is the electric repulsion greater than the gravitational attraction? FE / FG = 4 x 1042 • By what factor is the electric force between two protons greater than the gravitational force between them. FE / FG = 1.2 x 1036 • Given the difference in magnitudes of gravitational and electrical forces, why do you feel gravitational attraction from the earth, but no electrical forces? Electrically neutral • Human beings are electrically neutral objects to a high degree of accuracy. In this question you will estimate the force that would exist between 2 students standing one metre apart if they had just 1% of the electrons in their body removed, leaving them both positively charged. Take the mass of each student to be 60 kg and as a rough estimate, assume that humans are 100 % water. The molar mass of H20 (the mass of 6.02 x 1023 molecules) is 18 g. • How many water molecules do the students contain? • How many electrons are there in a water molecule? • How many electrons are there in total in each student? • Taking 1 % of these away will leave each student with a net positive charge of 1% of their electrons. What is its value? • Now calculate the force between the two students, if they are standing 1 metre apart and comment.

16. How many water molecules do the students contain? (60/0.018) X 6.02 X 1012 = 2.0 X 1027 • How many electrons are there in a water molecule? 10 (8 from O and 1 from each H) • How many electrons are there in total in each student? 2.0 x 1028 • Taking 1 % of these away will leave each student with a net positive charge of 1% of their electrons. What is its value? 1 % = 2.0 x 1026 therefore charge = 2.0 x 1026 x 1.6 x 10-19 = 32 MC • Now calculate the force between the two students, if they are standing 1 metre apart and comment. F = 9.3 x 1024 N This is a huge force on each student, almost the weight of our planet!

17. Electric Field Strength for a Point Charge • Consider the force felt by a charge q in the field of another charge Q, where the charges are separated by a distance r. F = kQq / r2 • But E = F/q, so E = k Q / r2 • The electric field strength for a point charge Q is inversely proportional to the square of the distance from the point charge. +Q +q F r

18. Worked Example Work out the field strengths at the points labelled A and B in the diagram below. What do you notice about the values, and why is this? Add arrows at A and B to indicate the electric field strengths there. 10 cm -5C 20 cm B A EA = kQ/r2 = -1.124 x 1012 N C-1 EB = kQ/r2 = -4.5 x 1012 N C-1 Values are negative because the charges are negative indicating an attractive force.

19. Electric Field Strength Between Charged Parallel Plates + + + + + + ++ + + + + + + + - - - - - - - - - - - E = V/d • For a given separation and p.d, the electric field strength between two parallel plates is constant (uniform field), except at the edges. A point charge q is moved between the plates and experiences a constant force F due to the uniform electric field. • Work done on the charge = energy transformed by the charge Fd = Vq • Rearranging this equation, we have F / q = V / d • By definition, F/q = E, so E = V / d • E is directly proportional to the p.d between the two plates and inversely proportional to the separation of the plates. • Units of electric field strength are V m-1 or N C-1 +q d

20. Worked Example A charged dust particle is stationary between two horizontal charged metal plates. The metal plates have a separation of 3.6 cm and the p.d between the plates is 720 V. The dust particle has a charge of +7e, where e = 1.6 x 10-19 C. Calculate: • The electric field strength between the plates E = V / d = 720 / 0.036 = 2.0 x 104 V m-1 • The weight of the dust particle The particle is stationary, therefore the net force must be zero. Weight = Electric force = Eq = 2.0 x 104 x (7 x 1.6 x 10-19) = 2.2 x 10-14 N

21. To summarise