SACE Stage 1 Physics Electric Fields

Introduction Consider two charges, the force between the two charged bodies is inversely proportional to the square of the distance between them

Introduction The force between two charges is directly proportional to the strength of each charge. If q 1 and q 2 represent the strength of the charges on each of the two spheres then,

Introduction This leads to the conclusion that the force is directly proportional to the product of the two charges.

Introduction Combining these results, we can conclude that the force between two charges is proportional to the product of the two charges and inversely proportional to the square of the distance between the two charges.

The Unit of Charge Charges are measured in Coulombs (C). Typically charged bodies are of the order of  C (10 -6 C). The charge on an electron is negative and is usually given the symbol e, and e = 1.6 x C.

The Unit of Charge The charge on a proton is positive and equal in magnitude to e. It takes 6.25 x electrons or protons to create a charge of one coulomb.

Coulomb’s Law The magnitude of the electrostatic force between the two charges is directly proportional to the magnitudes of the charges and inversely proportional to the square of the distance between the charges.

Coulomb’s Law q 1 and q 2 are the magnitudes of the charges on the two bodies. r is the distance between them. k is the constant of proportionality, k = 9 x 10 9 Nm 2 C -2.

Coulomb’s Law To determine if the force is attractive or repulsive can be determined as follows, Any two stationary charges experience mutual forces along the line joining the two centres of the charges. The forces are attractive if the charges are of opposite sign and repulsive if the two charges are of the same sign.

Coulomb’s Law In both cases, F 1 = -F 2. That is the forces are of equal magnitude but opposite in direction (Newton’s 3 rd Law).

Dependence of Force on the Medium Between the Charges Coulomb’s Law, replacing k,

Dependence of Force on the Medium Between the Charges The quantity  0 is called the permitivity of a vacuum. Its value is 8.85 x C 2 N -1 m -2. If the medium between the two charges is replaced with oil, plastic or some other insulating medium then the value of  0 will need to change.

How big a charge is one Coulomb? 1 Coulomb is very large. Consider 2 charges of 1C each, separated by a distance of 1m in air.

How big a charge is one Coulomb? Consider 2 charges of 1C each, separated by a distance of 1km in air. The weight of a small car!

Examples Calculate the force of repulsion between two small spheres each carrying a charge of 4.0  C, 0.5m apart in a vacuum.

Examples The magnitude of the force between two charges is 5.00N. What is the magnitude of the force if, (a)One charge is doubled in magnitude (b)The magnitude of one charge is multiplied by 3 and the other is multiplied by ¼. (c)The distance between the charges doubles.

Examples (a)One charge is doubled in magnitude q 1 is multiplied by 2  F is multiplied by 2,  F = 10.0N (b)The magnitude of one charge is multiplied by 3 and the other is multiplied by ¼. q 1 x 3 and q 2 x ¼  F is multiplied by 3 and by ¼, ie ¾.  F = ¾ x 5N ie F = 3.75N

Examples (a)The distance between the charges doubles. If r is multiplied by 2 then F is multiplied by ( 1 / 2 ) 2, ie by a ¼.  F = ¼ x 5N ie F = 1.25N

Charges in a Straight Line (1D) If you have 3 charged spheres A, B and C and you wish to find the force on Charge C due to the other 2 charges you would, Use Coulomb’s Law to find the force on C due to A. Then repeat to find the force on C due to B. The add the two forces vectorially to find the resultant force on C (F C = F A + F B ).

Charges in a Straight Line (1D) If you have 3 charged spheres A, B and C and you wish to find the force on Charge C due to the other 2 charges you would, Use Coulomb’s Law to find the force on C due to A. Then repeat to find the force on C due to B. Then add the two forces vectorially to find the resultant force on C (F C = F A + F B ).

Charges in a Straight Line (1D) Two Charges, q 1 and q 2, of magnitude 20  C and 10  C respectively, are placed 50cm apart in a vacuum. A third charge, q 3 = +30  C, is placed midway between q 1 and q 2. Find the force on q 3 if, 1.q 1 and q 2 are positive, 2.q 1 is positive and q 2 is negative.

Charges in a Straight Line (1D) Firstly determine the magnitude of the force on q3 due to q 1 and q 2. q 1 = 20  C = 2.0 x C, q 2 = 10  C = 1.0 x C q 3 = 30  C = 3.0 x C, r = 25cm = 0.25m

Charges in a Straight Line (1D) Magnitude of force between q 1 and q 3,

Charges in a Straight Line (1D) Magnitude of force between q 2 and q 3,

Charges in a Straight Line (1D) (1)q 1 and q 2 are both positive F 13 is to the right and F 23 is to the left.

Charges in a Straight Line (1D) (1)q 1 is positive and q 2 is negative F 13 and F 23 are in the same direction

Charges in a Plane (2D) Two charges q 1 and q 2 of magnitude +2  C and -1  C respectively are placed 50cm apart in a vacuum. A third charge, q 3 = +3  C, is placed 40cm from q 1 so as to form a right angled triangle as in the direction of the diagram. Find the force on q 3.

Charges in a Plane (2D)

Due to Pythagoras’ Theorem, the distance from q 2 to q 3 is 30cm (a triangle).

Charges in a Plane (2D)

By Pythagoras’ Theorem,

Charges in a Plane (2D) Thus the resultant force on q 3 is 0.45N at an angle of 41.6 o above the horizontal.

The Electric Field An electric field exists in a region of space if a charged body, placed in that region, experiences a force because of its charge.

The Electric Field The direction of the electric field is determined by the direction of the field placed on a test charge in that field

Pictorial representation of the Electric Field Direction of Lines of Force Electric fields are represented by using lines of force. Lines of force show the direction and magnitude of the field strength at any point in the field.

Pictorial representation of the Electric Field Density of the Lines of Force The number of Fields Lines per unit of cross sectional area is proportional to the field strength. If the lines of force are close together, the electric field is stronger.

Pictorial representation of the Electric Field Drawing Lines of Force 1.Lines of force are always drawn away from positive charges and towards negative charges. 2.Lines of Force always leave or come into contact at right angles to the surface. 3.Lines of Force never cross each other.

Pictorial representation of the Electric Field Examples of Electric Fields An isolated Point Charge A positive chargeA negative charge

Pictorial representation of the Electric Field Examples of Electric Fields An Electric Dipole – Two Point Charges A positive and negative charge Two positive point charges.

Pictorial representation of the Electric Field Examples of Electric Fields A Charged Hollow Sphere A positively charged sphere A negatively charged sphere

Pictorial representation of the Electric Field Examples of Electric Fields Two Oppositely Charged Parallel Plates Notice that the field is curved at the ends.

Pictorial representation of the Electric Field Examples of Electric Fields A Non-Uniform Conductor The smaller the radius of curvature of a surface, the greater the concentration of charge in that region.

Pictorial representation of the Electric Field Corona Discharge If the intensity of the electric field near a sharp projection of a conductor can be large enough, charge can leak away from this point. The gradual discharge is called corona discharge.

Electric Field Strength (E) Definition The electric field strength vector E at any point in an electric field is defined as the force per unit positive (test) charge placed at that point in the field. Units are NC -1

Electric Field Strength (E) To find the electric field strength at a point A, in the field of an isolated point charge, +Q and at a distance r from it, as in the diagram at the right, place a small positive (test) charge +q, at the point A.

Electric Field Strength (E) The force between the 2 charges, Electric field strength at A is given by,

Electric Field Strength (E) Example, 1.Calculate the electric field strength E at a point X, 3cm from a charge of 6  C in a vacuum.

Electric Field Strength (E) Example, 1.Calculate the electric field strength E at a point X, 3cm from a charge of 6  C in a vacuum.

Electric Field Strength (E) Example, 2.If a charge of C is placed at X, what force would it experience?

Electric Field Strength (E) Example, 2.If a charge of C is placed at X, what force would it experience?

Electric Field Strength (E) Example, 3.What other information would you need if you were to calculate the acceleration of the masses due to the Coulombic force?

Electric Field Strength (E) Example, 3.What other information would you need if you were to calculate the acceleration of the masses due to the Coulombic force? Would need to know the masses of the bodies on which the charges reside as,

Electric Field Strength (E) Example, 4.(a)What is the electric field strength 6cm from the 6  C charge?

Electric Field Strength (E) Example, 4.(a)What is the electric field strength 6cm from the 6  C charge? As E  1 / d 2, if d is doubled to 6cm, then E is reduced to ¼ of its original value. E = 1.5 x 10 7 NC -1, in a direction away from the 6  C charge.

Electric Field Strength (E) Example, 4.(b)What is the electric field strength 1cm from the 6  C charge?

Electric Field Strength (E) Example, 4.(b)What is the electric field strength 1cm from the 6  C charge? If d is reduce to1cm ( 1 / 3 its original value) then E is multiplied by a factor of 9. E = 5.4x10 8 NC -1, in a direction away from the 6  C charge.

Superposition of Electric Fields We can calculate the electric field strength due a point charge using, If we want to find the electric field strength at a point due to one or more charges, we must calculate the individual field strengths due to each charge and then add them vectorially.

Superposition of Electric Fields Two Charges of +2.0  C and -3  C are separated by a distance of 20cm in a vacuum. (1)Calculate the resultant electric field strength at X. (2)What would be the electric field strength at X if the - 3  C charge were changed to +3  C?

Superposition of Electric Fields (1) Calculate the resultant electric field strength at X. q 1 = +2  C = 2 x 10-6 Cq 2 = -3  C = -3 x 10-6 C

Superposition of Electric Fields q 1 = +2  C = 2 x 10-6 Cq 2 = -3  C = -3 x 10-6 C The charge due to q 2 is 1.5 times bigger than q 1 due to proportionality.

Superposition of Electric Fields (b)What would be the electric field strength at X if the - 3  C charge were changed to +3  C? If q 2 is made positive then E 2 is directed to the left.

Superposition of Electric Fields Example – Two dimensional superposition. Two charges q 1 = +2  C and q 2 = -3  C are separated by a distance of 50cm in a vacuum. Calculate the electric field strength at the point Y, which is 40cm from q  and 30cm from q 2.

Superposition of Electric Fields q 1 = +2  C = 2 x C,q 2 = -3  C = -3 x C Electric Field strength due to q1, Electric Field strength due to q2, By Pythagoras’ Thm,

Electric Potential Electric Potential Energy Any charge placed in an electric field has energy due to its position in the field. We call this energy – electric potential.

Electric Potential Electric Potential Difference Definition – The electric potential difference between two points in an electric field is the work done per unit charge in moving a positive (test) charge between the two points, provided all other charges involved remain undisturbed.

Electric Potential The unit of electric potential difference Work is measured in Joules and charge is measured in Coulombs, therefore the unit is defined as,

Electric Potential Example How much work is done when moving an electron across a potential difference of 100 volts?

Electric Potential Energy Changes Whenever a charge q moves between 2 points in an electric field, with potential difference  V, its potential energy changes according to the relation,

Electric Potential The law of conservation of energy enables us to use the one expression (q  V) to determine, The work done when a charge q moves through a potential difference The change (increase or decrease) in electric potential energy  PE. The change (decrease or increase) in kinetic energy  K.

The Electron Volt as a Unit of Energy Definition One electron volt (1eV) is the work done when an electron moves through a potential difference of one volt. Or One electron volt (1eV) is the energy gained or lost by an electron in moving through a potential difference of one volt.

The Electron Volt as a Unit of Energy Consider the following, If an electron - charge = -e, moves through a potential difference of 300V, the change in potential energy is 300eV. If a proton - charge = +e, moves through a potential difference of 2500V, the work done is 2500eV

The Electron Volt as a Unit of Energy If an  -particle - charge = +2e, moves through a potential difference of 400V, the change in potential energy is 800eV. If a nitrogen ion - charge = -3e, moves through a potential difference of 50V, the change in potential energy is 150eV.

The Electron Volt as a Unit of Energy Conversion between electron volts and joules. An electron moves through a potential difference of 1 Volt. The work done is 1 electron Volt.

The Electron Volt as a Unit of Energy Example In an X-ray tube electrons are accelerated across a potential difference of 30,000V. (1)What energy do they gain? (2)What is their final kinetic energy?

The Electron Volt as a Unit of Energy (1)What energy do they gain?

The Electron Volt as a Unit of Energy (2)What is their final kinetic energy? This potential energy has been converted to kinetic energy. Thus the final kinetic energy of the electron is,