Electrostatics

Learning Objectives The electrostatic force (Coulomb’s Law) can be either repulsive or attractive (SOL 12.a) The interaction of two particles can be described as: the creation of a field by one of the particles and the interaction of the field with the second particle (SOL 12.b).

Magnitude of charge on protons and electrons are exactly the same Protons have a positive charge Electrons have a negative charge Neutral atoms contain equal numbers of protons and electrons

Insulators and Conductors Need to know Insulator: electrons are bound very tightly to the nuclei. Wood and rubber are good insulators. Conductor: electrons are bound very loosely and can move about freely. They are often referred to free electrons. Metals are good conductors. Semiconductor: very few free electrons (silicon, germanium and carbon)

Static Electricity You have probably experienced a charge lately (comb, dryer, carpet, car seat, …) An object becomes charged due to a rubbing process and is said to possess a net electric charge An item containing a net positive charge has lost electrons An item containing a net negative charge has gained electrons

Law of Conservation of Electric Charge Need to know The net amount of electric charge produced in any process is zero If one object or one region of space acquires a positive charge, then an equal amount of negative charge will be found in neighboring areas or objects

Unlike Charges Attract; Like Charges Repel Need to know

3 Ways to Charge an Object Need to know Friction: Rubbing two objects together with different electron attachment. Heat generated frees electrons to join object with stronger attachment.

3 Ways to Charge an Object Need to know 2. Conduction: Electrons are transferred from one object to another by touching. Usually it involves moving from one electric potential to another. John TraVOLTa Demo

3 Ways to Charge an Object Need to know 3. Induction: Rod does not touch sphere. It pushes electrons out of the back side of the sphere and down the wire to ground. The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed.

Electromagnetism One of the four fundamental forces of the universe (electromagnetism, gravity, weak nuclear and strong nuclear forces) The forces that act between atoms and molecules to hold them together are electrical forces Elastic, normal and contact forces (pushes and pulls) result from electric forces acting at the atomic level

Forces resulting from charges Charges push and pull on one another Closer the charge the higher the force The stronger the charge the higher the force

Coulomb’s Law Need to know The magnitude of the force between charge qA and charge qB, separated a distance d, is proportional to the magnitude of the charge and inversely proportional to the square of the distance: F = K qAqB d2 qA qB d

Coulomb’s Law: Key Facts Need to know The charge of an electron is: -1.60 x 10-19 coulombs (C) The charge of a proton is: 1.60 x 10-19 coulombs (C) The charge, q, is measured in coulombs. The distance, d, is measured in meters. The force, F, is measured in newtons. The constant, K = 9.0 x 109 Nm2/C2

Problem Solving Strategy Sketch the system showing all distances Diagram the vectors Use Coulomb’s law to find the magnitude of the force. Note: it is unnecessary to include the sign of the charges or the distance. The answer is always positive. Use your diagram along with trigonometric relations to find the direction of the force

Example Problem 1 Two charges are separated by 3.0 cm. Object A has a charge of +6.0 C, while object B has a charge of +3.0 C. What is the force on object A? Known: Unknown: qA = +6.0 x 10-6 C FB on A = ? qB = +3.0 x 10-6 C d = 0.030 m

Example 1 Solution F = K qAqB d2 = (9.0 x 109 Nm2/C2)(6.0 x 10-6C)(3.0 x 10-6C) (3.0 x 10-2 m)2 FB on A = 1.8 x 102 N

Example 2: Three Charges Given: Find the net force on the -2 µC charge Known: -2 µC +6 µC 2 µC 6 cm 2 cm

FNet = FA on B + FC on B = - 30 N + 90 N = 60 N +6 µC FA on B= KqAqB d2 = (9x109 Nm2C2)(6x10-6 C)(2x10-6 C) (0.06 m)2 = - 30 N FC on B= KqCqB = (9x109 Nm2C2)(2x10-6 C)(2x10-6 C) (0.02 m)2 = + 90 FNet = FA on B + FC on B = - 30 N + 90 N = 60 N -2 µC +6 µC 2 µC 6 cm 2 cm

Example Problem 3 A sphere with a charge 6.0 C is located near two other charged spheres. A -3.0 C is located 4.00 cm to the right and a 1.5 C sphere is located 3.00 cm directly underneath. Determine the net force on the 6.0 C sphere. FB on A FC on A Fnet  A qA= 6 C qc = 1.5 C qB = -3 C C B dAB dAC

Example 3 Solution FB on A = FC on A = Fnet =  = B A dAB Fnet qA= 6 C qc = 1.5 C qB = -3 C C B dAB dAC FB on A FC on A Fnet  FB on A = FC on A = Fnet =  =

Static Charge Generator

Electric Field Need to know An electric field extends outward from every charge and permeates all of space

Investigating the Electric Field We can quantify the strength of an electric field by measuring the force on a small positive test charge So small that the force it exerts does not significantly alter the distribution of the charges that create the field qA+ +qB a

Electric Field An electric field, E, at any point is defined as the force, F, exerted on a tiny positive test charge at that point divided by the magnitude of the test charge: E = F/qB qA+ +qB

Electric Field Equation E = F/qB E = K qB qA/r2 qB E = KqA/r2 qA+ +qB

Example Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge qA = -3.0 x 10-6 C E = KqA/r2 Answer: 3.0 x 105 N/C

Electric Field Lines Drawn so that they indicate the direction of the force due to the given field on a positive charge qA+ +qB

Electric Field Lines Need to Know Lines indicate direction of the force due to the given field on a positive test charge

Properties of Field Lines Need to know The field lines indicate the direction of the electric field The lines are drawn so that the magnitude of the electric field, E, is proportional to the number of lines crossing unit area perpendicular to the lines. The closer the lines, the stronger the field. Electric field lines start on positive charges and end on negative charges

Electric Potential Difference Need to know V = Won q’ = PE: Potential difference often q’ q’ referred to as Voltage Electric Potential Difference Units: Volt =J/C + g E displacement displacement W = Fd = mgd W = Vq + Big Negative Charge

Typical Voltages Source Voltage Thundercloud to ground High voltage power line Power supply for TV tube Auto ignition Household outlet Auto battery Resting potential across nerve membrane Potential changes on skin (EKG) Voltage 108 V 106 V 104 V 102 V 12 V 10-1 V 10-4 V

Example Two parallel plates are charged to a voltage of 50V. If the separation between the plates is 0.050 m, calculate the electric field between them. E = V/d = 50V/0.050m =1000V/m + - E = 1000 V/m d = 5 cm

Practice Two parallel plates are charged to a voltage of 500V. If the separation between the plates is .0050 m, calculate the electric field between them. Two parallel plates are charged to a voltage of 50V. If the electric field between them is 10,000 V/m, what is distance between the plates? Answers: 100,000 V/m .0050 m

Capacitors Need to Know A capacitor is a device that can store electric charge Consists of two conducting objects placed near each other but not touching They store charge for later use Usage: camera flash, energy back-up for computers and as surge protectors

Capacitors Consists of a pair of parallel plates of area, A, and separated by a small distance d. In a diagram, they are represented by the symbol: If a voltage is applied to a capacitor, one plate acquires a negative charge and the other an equal amount of positive charge.

Capacitors Need to Know The amount of charge acquired by each plate is proportional to the potential difference Q = CV Where C is constant and is called the capacitance of the capacitor Unit: Coulombs/Volt = Farad Typical capacitor range is 1pF (10-12) to 1F (10-6)

Determining Capacitance Constant for a given a capacitor Depends on structure and dimensions of he capacitor itself: C = o A/d A = area d = separation distance between plates o = 8.85 x 10-12 C2/Nm2 = permittivity of free space