Electrostatics Learning Outcomes for the Lesson: state Coulomb’s Law Use Coulomb’s Law to solve problems Define Electric Field -calculate net Electric field for two point charges Describe electric field for simple charge distributions in uniform and non-uniform situations Solve problems for electric field Use Electric Potential Energy equations to solve problems 13/04/2019

Electrostatics Vocabulary Electric Force Electric Charge Electric Field Electric Potential Energy Electric Potential (Voltage) electromotive force (emf), potential difference, Potential Current Coulomb Ampere Conductor Coulomb’s Constant Coulomb’s Law 13/04/2019

Electric Force Electric force is fundamental to our very existence! All atoms and molecules are held together by this electric force. Without the electric force there would be no such things as molecules (NO DNA & NO LIVING THINGS!) 13/04/2019

Electric Force We can move because of this force: (Muscle require chemical reactions to operate.) All forms of transportation require the use of this force. This is important Stuff! 13/04/2019

Electrostatic Force Law of Electrostatic Charges Like Charges repel. Unlike Charges attract. Charged objects attract neutral objects 13/04/2019

Electrostatic Force + + + + + + + + Attraction - - - - - - - - - - + + + + + + + + - - - - - - - - - - Attraction Charge is built up by rubbing two objects together 13/04/2019

Electric Charge Atoms are made of three sub-atomic particles: protons, electrons, and neutrons Protons are found in the nucleus and have a positive charge Electrons are found outside the nucleus and have a negative charge Neutrons are found in the nucleus and carry no charge 13/04/2019

Electric Charge The simplest possible charge called the elementary charge is the charge on an electron (-) or proton (+) It is given a value of e = 1.6 x 10-19 C Like charges repel and unlike charges attract (that is why electrons and protons are attracted to each other) 1 Coulomb (C) of charge is equal to the number of electrons that flow past a conductor in 1 second when 1 ampere of current flows (1 C = 6.25 x 1018 charges) 13/04/2019

Electrostatic Force Electric Charge is always conserved. When one object receives a positive charge another object must receive an equal but opposite negative charge. The charges are due to the sub-atomic particles within an atom Proton (+) Electron (-) 13/04/2019

Electrostatic Force Coulomb’s Law: kQq Fe = r2 Fe is the electric force (N) Q is the charge of one object (C) q is the charge of the other object in (C) r = the distance between the two objects (m) k is Coulomb’s Constant = 9.00 x 109 Nm2/C2 13/04/2019

Coulomb’s Law 24 μC -15 μC 2.3 m What is the electrostatic force between the two charges? 9.00 x 109 x 24 x 10-6 x -15 x 10-6 2.32 Fe = k Q q r2 = Fe = -0.61 N (Attractive) 13/04/2019

Coulomb’s Law 22 μC -13 μC 2.3 m 45 μC 1.7 m What is the net force on the -13 μC due to the other two charges? Solution: Find the force on each side of the center charge. Fel = k Qq r2 Fer = kQq r2 13/04/2019

Coulomb’s Law Fe(left) = 9.00 x 109 x 22 x 10-6 x -13 x 10-6 1.72 = 0.891 N (Left) Fe(right) = 9.00 x 109 x 45 x 10-6 x -13 x 10-6 2.32 = 0.995 N (right) Therefore: Fnet = 0.995 N (R) - 0.891 N (L) = 0.104 N (R) 13/04/2019

Coulomb’s Law vs Gravitational Law Fe = k x Q x q r2 Fg = G x M x m r2 Note the similarity of the two graphs FE r Fg r 13/04/2019

Coulomb’s Law vs Gravitational Law FE 1/r2 Fg 1/r2 m = KQq m = GMm Note the similarity between the two laws. Scientists have been trying to link the laws together for hundreds of years. So far … no connection exists! 13/04/2019

Electric Field Electric Field: Remember when we talked about gravitational field to describe how forces act over a distance… well the same thing applies to electric force. Electric Field: Shows the direction that a positive-point charged would travel near another charge. Shown as vector arrows. Electric field lines point away from a positive charge. 13/04/2019

Electric Field Electric Field around a negative charge Electric Field around two equal and opposite charges 13/04/2019

Electric Field Electric Field around two equal charges Electric Field around various unequal charges 13/04/2019

Electric Field Electric Field between two charged plates + - Electric Field between two charged plates The Electric field is constant between the plates (except at the ends). 13/04/2019

Electric Field If two metal plates (called a capacitor) are oppositely charged by a power supply a uniform electric field is created between the plates + - P.S. 13/04/2019

Electric Field - Formula Electric Field strength is analogous to gravitational field strength. (Remember: g = F/m and g = GM/r2) By definition, electric field equals force per charge: E = F q F = k x Q x q r2 and Putting these two together results in: 13/04/2019

Electric Field - Formula k x Q r2 E is the electric field intensity (N/C) Q is the charge of one object (C) r = the distance between the two objects (m) k is Coulomb’s Constant = 9.00 x 109 Nm2/C2 13/04/2019

Electric Field - Calculations P 1.5 m 65 μC What is the electric field intensity at a distance of 1.5 m away from a 65 μC charge? E = k x Q r2 New Equation E = 9.00 x 109 x 65 x 10-6 1.52 E = 260000 N/C 13/04/2019

Electric Field - Calculations EB P 45 μC -15 μC 3.4 m 2.8 m ER What is the net electric field at point P between the two oppositely charged particles? Solution: Find the E-fields due each charge separately ER = k x Q r2 EB = k x Q r2 13/04/2019

Electric Field - Calculations ER = 9.00 x 109 x 45 x 10-6 3.42 EB = 9.00 x 109 x -15 x 10-6 2.82 ER = 35034 N/C (Right) EB = 17219 N/C (Right) Enet = 52000 N/C (Right) 13/04/2019

Electric Potential Energy (Non-Uniform E Fields) We are now going to look at how electric potential energy and electric force is related in non-uniform e- fields (outside of charged plates) We have to generate some new equations because of this new situation. 13/04/2019

Electric Potential Energy Remember W = ΔEk And W = F x d Therefore: ΔEk = ΔEp = F x d 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) W = ΔEp = F x d F = KQ1Q2 R2 But: Therefore: KQ1Q2 Ep = R2 x d 13/04/2019

Electric Potential Energy (Non-Uniform E Fields) Since d = R, one R will cancel, leaving: KQ1Q2 Ep = R Note: This is very similar to Gravitational Potential Energy: - GM1m2 Ep = R 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Example #1: What is the potential energy of the –2.6 μC charge relative to the +37 μC Charge? 7.3 m +37 μC -2.6 μC 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Solution: KQ1Q2 Ep = R 9.00 x 109 x 37 x 10-6 x –2.6 x 10-6 Ep = 7.3 Ep = - 0.119 J 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Example #2 (A more realistic problem is when one charge is forced to move or is allowed to move on its own.) 2.4 m 0.75 m 4.0 μC 2.0 μC What work is done in moving the 2.0 μC charge 0.75 m closer to the 4.0 μC charge? 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Solution: Work = ΔEp ΔEp = Ep2 - Ep1 kQq r2 kQq r1 ΔEp = 9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6 1.65 9.00 x109 x 4.0 x 10-6 x 2.0 x 10-6 2.4 ΔEp = ΔEp = 0.0436 J - 0.030 J = 0.0136 J 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Now Try this: An electron travelling at 6.0 x 105 m/s when it is heading straight for this iron atom. What is the distance for the electron’s closest approach? Fe atom -26 elementary Charges v = 6.0 x 105 m/s d = ? electron 13/04/2019

Electric Potential & Energy (Non-Uniform E Fields) Solution: Ek = Ep KQ1Q2 R ½ m v2 = ½ x 9.11 x 10-31 x (6.0 x 105)2 = 9.00 x 109 x 26 x - 1.6 x 10 –19 x - 1.6 x 10-19 R R = 5.99 x 10-27 1.64 x 10-19 R = 3.7 x 10-8 m (about the size of an atom!!) 13/04/2019

Electric Potential (Voltage) Not to be Confused with electric potential energy! Other words for the same term: Voltage, electromotive force (emf), potential difference, Potential The “driving force” that causes electrons (and other charges) to move from one place to another “electrical pressure” 13/04/2019

Electric Potential Relates the potential energy (or work) done on an object and the amount of charge on that object Book Definition: Electric potential equals potential energy per unit charge. For example: a 1.5 V dry cell can give 1.5 Joule of energy per coulomb of charge moved. 120V circuit can give 120 J of energy per coulomb of charge 13/04/2019

Electric Potential Fg mg FA A B m Epi Epf Gravity Field FE qE FA B A q + Applied Force + Electric Field Fg mg FA A B m Epi Epf Gravity Field An object, with mass m, being raised in a uniform downward gravitational field by a force, F, experiences a downward gravitational force Fg = mg. FE qE FA B A q Work must be done in moving a charge against an electric field. Epi Epf ElectricField An object, with charge q , being raised in a uniform downward electric field by a force, F, experiences a downward electric force FE = qE.

Electric Potential & Energy charge V = Ep Q (Units = J/C = Volt) Note: Ep = QV is a very useful equation to know! And Ep = KQq R

Electric Potential & Energy (Non-Uniform E Fields) Therefore: Electric Potential or Voltage at a Point: KQq R q V = Therefore: V = KQ R Electric potential for a non-uniform E-field situation

Electric Potential (Non-Uniform E Fields) Q Electric Field Equipotentials for a small positive charge: V3 = 100 V V4 = 50 V V1 = 200 V V2 = 150 V V = kQ R V1 = 200 V The same V exist anywhere along each of the equipotential concentric circles. V2= 150V V4 = 50 V V3 = 100 V 13/04/2019

Electric Potential (Non-Uniform E Fields) Example #1: 10.3 m P What is the electric potential at point P due to the 32 μC charge? 32 μC

Electric Potential (Non-Uniform E Fields) Solution: V = KQ R V = 9.00 x 109 x 32 x 10-6 10.3 m V = +28000 V 13/04/2019

Electric Potential (Non-Uniform E Fields) Example #2: 3.2 m 10 μC P 3 μC What is the electric potential at point P midway between the two charges? 13/04/2019

Electric Potential (Non-Uniform E Fields) Solution: Calculate the Voltage (electric potential) due to each charge separately and then just add them. VR = kQ r VL = kQ r VR = 9.00 x109 x 10 x 10-6 1.6 VL = 9.00 x109 x 3 x 10-6 1.6 VR = 56250 V VL = 16875 V V(at point P) = 73125 V 13/04/2019

Electric Potential (Non-Uniform E Fields) For a real challenge try this: What is the potential difference between points A and B shown below? 25 uC A B 3.0 m 1.7 m 13/04/2019

Electric Potential (Non-Uniform E Fields) Solution: Find the Voltage at each point due to each charge then find the voltage difference. At Point A: VA1 = kQ r VA2 = kQ r VA1 = 9.00 x109 x 25 x 10-6 3.0 VA2 = 9.00 x109 x 25 x 10-6 3.0 VA1 = 75000 V VA2 = 75000 V VA = 150000 V

Electric Potential (Non-Uniform E Fields) At Point B: VB1 = kQ r VB2 = kQ r VB1 = 9.00 x109 x 25 x 10-6 1.7 VB2 = 9.00 x109 x 25 x 10-6 1.7 VB1 = 132352 V VB2 = 132352 V VB = 264704 V 13/04/2019

Electric Potential (Non-Uniform E Fields) Therefore the Voltage difference is ΔV = VB - VA ΔV = 264704 V - 150000 V ΔV = 114704 V 13/04/2019

Electric Potential & Energy for uniform Electric Field Situations + - P.S. Uniform E Field Found in between two charged plates or a capacitor. 13/04/2019

Electric Potential & Energy for uniform Electric Field Situations V = Ep Q (Units = J/C = Volt) Re-arranging this equation results in a very useful form: Ep = Q x V 13/04/2019

Electric Potential & Energy Examine the following cathode-anode arrangements: Which has the greatest electric potential and hence the greatest energy? +200 V 0 V -200 V -400 V -600 V a) b) c) 13/04/2019

Electric Potential & Energy Example #1: What is the impact speed of electrons on the red plate when accelerated by the uniform electric field shown? 350 V v1 = 0 v2 = ? Ep Ek Ep = QV Ek= ½ mv2 But Ep is completely transferred to Ek! Ep = Ek QV = ½ mv2 13/04/2019

Electric Potential & Energy Therefore: 350 V v1 = 0 v2 = ? 2QV m v2 = v = 2 x 1.6 x 10-19 x 350 9.11 x 10-31 v = 1.1 x 107 m/s 13/04/2019

Electric Potential & Energy Example #2: What is the impact speed of the electron in this case? v2 = ? 125 V v1 = 2.0 x 106 m/s Ek1 Ep1 = QV Ek2 Try: Ek2 = Ek1 + Ep1 (Ans: 6.9 x 106 m/s) 13/04/2019

Electric Potential & Electric Field (Uniform E Field) Let us take a look how these two ideas can be related: W = Ep = QV (Potential electric Energy) E = F q (Electric Field Equation) And therefore: F = q x E 13/04/2019

Electric Potential & Electric Field (Uniform E Field) But: W = F x d Therefore: W = q x E x d And: W = Ep = QV Putting both of these equations together: q x E x d = Q x V Now: q = Q and cancels out on both sides 13/04/2019

Electric Potential & Electric Field (Uniform E Field) This leaves you with a very useful equation that relates electric potential and electric field. V = E x d V = voltage across plates (V) E = electric field in the plates (N/C) or (V/m) d = distance between the plates (m)

Electric Potential & Electric Field (Uniform E Field) By a little re-arranging: Voltage across the plates E = V d Distance bewteen the plates This again a useful equation in solving problems inside CRT’s and other devices that have charge plates or capacitors. 13/04/2019

Electric Potential & Electric Field (Uniform E Field) For Example: 230 V 0.025 m 0.175 m a) What is the E-field (Magnitude & Direction in between the plates? b) What is the electric force on the electron while it is in the plates? c) What is the acceleration of these electrons? d) Assuming they start from rest what would the speed of these electrons just before they hit the positive plate?

KQq R2 KQq R V KQ R2 KQ R Summing up all the electrostatic Equations F Non- Uniform Field Situation x R F (Force) Ep (Energy) KQq R2 KQq R ÷ q (Electric Field) E V (Electric Potential) KQ R2 KQ R

qE QV qV/d V V V d Summing up all the Electrostatic Equations F Ep E Uniform Field Situation: x d F (Force) Ep (Energy) qE QV or qV/d ÷ q (Electric Field) E V (Electric Potential) V d V 13/04/2019