Electrostatics 1) electric charge: 2 types of electric charge: positive and negative 2) charging by friction: transfer of electrons from one object to another 3) positive object: lack of electrons negative object: excess of electrons 4) Types of materials: a) Conductors: materials in which electric charges move freely (e.g. metals, graphite) b) Insulators: materials in which electric charges do not move freely (e.g. plastic, rubber, dry wood, glass, ceramic) c) Semiconductors: materials with electrical properties between those of conductors and insulators (e.g. silicon) d) Superconductors: materials in which electrical charges move without resistance (e.g. some ceramics at very low temperatures) roperties of Atomic articles e = elementary unit of charge (magnitude of charge on electron) article Mass Electric Charge Electron roton m e = 9.110 x 10-31 kg m p = 1.673 x 10-27 kg q = -e q = -1.60 x 10-19 C q = +e q = +1.60 x 10-19 C e = 1.60 x 10-19 C Neutron m n = 1.675 x 10-27 kg q = 0 q = 0 C 1. A balloon has gained 2500 electrons after being rubbed with wool. What is the charge on the balloon? What is the charge on the wool? q = -4.0 x 10-16 C q = +4.0 x 10-16 C 2. A rubber rod acquires a charge of -4.5 µc. How many excess electrons does this represent? 2.8125 x 10 13 e Conservation of Electric Charge: The total electric charge of an isolated system remains constant. 1
Electric Force (Electrostatic Force, Coulomb Force) Coulomb s Law: The electric force between two point charges is directly proportional to the product of the two charges and inversely proportional to square of the distance between them, and directed along the line joining the two charges. Coulomb Force F e qq =k r 1 2 2 k = Coulomb constant (electrostatic constant) k = 8.99 x 10 9 N m 2 C -2 NOTE: +-F denotes direction of force not sign of charge oint charge: a charged object that acts as if all its charge is concentrated at a single point Alternate formula for Coulomb force: F F e e = = 1 qq 4πε r 0 qq 4πε 1 2 2 0r 1 2 2 k = 1/ 4πε 0 ε 0 = permittivity of free space = 8.85 x 10-12 C 2 N -1 m -2 Use the Coulomb force to estimate the speed of the electron in a hydrogen atom. 2
The rinciple of Superposition The net electric force acting on a charged particle is the vector sum of all the electric forces acting on it. 1. Determine the net electrostatic force on charge q 1, as shown below. 2. Where can a third charge of +1.0 µc be placed so that the net force acting on it is zero? 3. Three point charges of -2.0 µc are arranged as shown. Determine the magnitude and direction of the net force on charge q 1. D = 2/3 m 3
Electric Field Electric field: a region in space surrounding a charged object in which a second charged object experiences an electric force Test charge: a small positive charge used to test an electric field Electric Field Diagrams 1. ositively charged sphere 2. ositive point charge 3. Negative point charge Radial Field: field lines are extensions of radii 5. Two positive charges 6. Two negative charges 7. Two unlike charges 8. Oppositely charged parallel plates roperties of Electric Field Lines Uniform Field: field has same intensity at all spots Edge Effect: bowing of field lines at edges 1. Never cross 2. Show the direction of force on a small positive test charge 3. Out of positive, into negative 4. Direction of electric field is tangent to the field lines 5. Density of field lines is proportional to field strength (density = intensity) 6. erpendicular to surface 7. Most intense near sharp points 4
Electric Field Strength Electric Field Strength (Intensity): electric force exerted per unit charge on a small positive test charge Electric Field: E F e = k 2 q Units: N/C Units: N Electric Force: F = Eq e Electric Field for a oint Charge: Qq Q 1 Q E= r = k = q r 4πε r 2 2 0 oint Charge Spherical Conductor 1. a) Find the magnitude and direction of the electric field at a spot 0.028 meter away from a sphere whose charge is +3.54 microcoulombs and whose radius is 0.60 centimeters. 2. a) Find the magnitude and direction of the gravitational field at an altitude of 100 km above the surface of the Earth. b) Find the magnitude and direction of the electric force acting on a -7.02 nc charge placed at this spot. b) Find the magnitude and direction of the gravitational force exerted on a 6.0 kg bowling ball placed at this spot. c) Find the electric field strength at the surface of the sphere. c) Find the gravitational field strength at the surface of the Earth. 5
3. a) Find the magnitude and direction of the net electric field halfway between the two charges shown below. b) Determine the electric force on a proton placed at this spot. 4. Two charged objects, A and B, each contribute as follows to the net electric field at point : E A = 3.00 N/C directed to the right, and E B = downward. What is the net electric field at? E = 3.61 N/C Theta =33.7 0 5. a) Two positive point charges, q 1 = +16 µc and q 2 = +4.0 µc, are separated in a vacuum by a distance of 3.0 m. Find the spot on the line between the charges where the net electric field is zero. 6
6. A proton is released from rest near the positive plate. The distance between the plates is 3.0 mm and the strength of the electric field is 4.0 x 10 3 N/C. a) Describe the motion of the proton. constant acceleration in a straight line b) Write an expression for the acceleration of the proton. c) Find the time it takes the proton to reach the negative plate. d) Find the speed of the proton when it reaches the negative plate. 7. A particle is shot with an initial speed through the two parallel plates as shown. a) Sketch and describe the path it will take if it is a proton, an electron, or a neutron. b) Which particle will experience a greater force? c) Which particle will experience a greater acceleration? d) Which particle will experience a greater displacement? 8. In the figure, an electron enters the lower left side of a parallel plate capacitor and exits at the upper right side. The initial speed of the electron is 5.50 10 6 m/s. The plates are 3.50 cm long and are separated by 0.450 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude. 7
High amount of E Electric otential Energy Gravitational otential Energy (E ) Reason for E : 1. Test object has mass (test mass = m) 2. Test mass is in a gravitational field (g) caused by larger object (M) Low amount of E 3. Larger object exerts a gravitational force on test mass (F g = mg) 4. Test mass has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Base level where E = 0 Gravitational potential energy: E = mgh W = E = mg h Reason for E : Electric otential Energy (E ) High amount of E 1. Test object has charge (test charge = +q) 2. Test charge is in an electric field caused by larger object (Q) 3. Larger object exerts an electric force on test charge (F E = Eq) Low amount of E 4. Test charge has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Base level where E = 0 Electric potential energy: E = Eq h W = E = Eq h Electric otential Energy (E )- the work done in bringing a small positive test charge in from infinity to that point in the electric field Derivation for oint Charges (Work done by field) E = W = Fs cosθ E = Eq s E E E r kq = ( q) ds 2 s kqq = s kqq = r r E = 0 Electric otential Energy due to a point charge Formula: Units: E kqq = r J Type: scalar 8
Electric otential (V) - work done per unit charge moving a small positive test charge in from infinity to a point in an electric field. Electric otential due to a point charge Formula: V V kqq E r q q = = kq Q = = r 4πε r 0 E = qv Units: J/C = volts(v) Type: scalar Higher potential B A Lower potential Zero potential Lower potential Higher potential Zero potential B A 1. a) Calculate the potential at a point 2.50 cm away from a +4.8 µc charge. b) How much potential energy will an electron have if it is at this spot? 3. What is the potential where a proton is placed 0.96 m from a -1.2 nc charge? 9
oint Charges +Q +Q +Q +Q Electric Force Electric Field Electric otential Energy Electric otential Two objects needed interaction between the two One object needed property of that one object Two objects needed quantity possessed by the system One object needed property of the field Magnitude: F = Eq Magnitude: E = F/q Magnitude: E = qv Magnitude: V = E /q F = kqq/r 2 E = kq/r 2 E = kqq/r V = kq/r Units: N Units: N/C Units: J Units: J/C Type: vector Type: vector Type: scalar (+/-) Type: scalar (+/-) Direction: likes repel, unlikes attract Direction: away from positive, towards negative Sign: use signs of Q and q Sign: use sign of Q Sign: don t use when calculating check frame of reference Sign: don t use when calculating check frame of reference F = 0 where E = 0 E = 0 where V = 0 10
1. a) Calculate the net electric field at each spot (A and B): b) Calculate the net electric force on a proton placed at each spot. 2. a) Calculate the net electric potential at each spot (A and B): b) Calculate the electric potential energy of a proton placed at each spot. 11
Electric otential and Conductors Graphs for a spherical conductor For a hollow or solid conductor, 1. all the charge resides on the outside surface 2. the electric field is zero everywhere within 3. the external electric field acts as if all the charge is concentrated at the center Electric Field Strength Value at surface = kq/r 2 Electric otential 4. the electric potential is constant ( 0) everywhere within and equal to the potential at the surface radius Distance radius Distance A spherical conducting surface whose radius is 0.75 m has a net charge of +4.8 µc. a) What is the electric field at the center of the sphere? b) What is the electric field at the surface of the sphere? c) What is the electric field at a distance of 0.75 m from the surface of the sphere? d) What is the electric potential at the surface of the sphere? e) What is the electric potential at the center of the sphere? f) What is the electric potential at a distance of 0.75 m from the surface of the sphere? 12
Equipotential Surfaces Equipotential surface: a surface on which the electric potential is the same everywhere 1. Locate points that are at the same electric potential around each of the point charges shown. 2. Sketch in the electric field lines for each point charge. 3. What is the relationship between the electric field lines and the equipotential surfaces? erpendicular Field lines point in direction of decreasing potential Electric otential Gradient The electric field strength is the negative of the electric potential gradient. Formula: Units: N/c or V/m V E= x For each electric field shown, sketch in equipotential surfaces. Sketch in equipotential surfaces for the two configurations of point charges below. http://wps.aw.com/aw_young_physics_11/0,8076,898593-,00.html http://www.surendranath.org/applets.html 13
Electric otential Difference Electric otential Difference ( V) work done per unit charge moving a small positive test charge between two points in an electric field Formula: Units: J/C = V W V = q V = E q E = q V High and Low otential 1. a) Which plate is at a higher electric potential? positive b) Which plate is at a lower electric potential? negative c) What is the electric potential of each plate? Arbitrary relative to base level d) What is the potential difference between the plates? Not arbitrary depends on charge, distance between, strength of electric field, geometry of plates, etc. e) Where will: a proton have the most electric potential energy? Mark plates with example potentials, as well as spots within field Mark ground mark equipotentials an electron? a neutron? an alpha particle? Not arbitrary 2. An electron is released from rest near the negative plate and allowed to accelerate until it hits the positive plate. The distance between the plates is 2.00 cm and the potential difference between them is 100. volts. a) Calculate how fast the electron strikes the positive plate. E o = E f E e = E K qv = ½ mv 2 v = sqrt (2qV/m) v = sqrt(2(1.6 x 10-19 )(100 V) /(9.11 x 10-31 )) v = 5.9 x 10 6 m/s b) Calculate the strength of the electric field. Formula: qv = ½ mv 2 Ve = ½ mv 2 Formula: E = - V/ x E = V/d 14
The Electronvolt Electronvolt: energy gained by an electron moving through a potential difference of one volt Derivation: E e = q V E e = (1e)(1 V) = 1 ev E e = (1.6 x 10-19 C)( 1 V) E e = 1.6 x 10-19 J Therefore: 1 ev = 1.60 x 10-19 J 1. How much energy is gained by a proton moving through a potential difference of 150. V? 150 ev or 150(1.60 x 10-19 ) = 2.4 x 10-17 J 2. A charged particle has 5.4 x 10-16 J of energy. How many electronvolts of energy is this? Factor-label (5.4 x 10-16 J) (1 ev/1.6 x 10-19 ) = 3375 ev 3. An electron gains 200 ev accelerating from rest in a uniform electric field of 150 N/C. Calculate the final speed of the electron. 4. In Rutherford s famous scattering experiments (which led to the planetary model of the atom), alpha particles were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.00 10 7 m/s directly toward the gold nucleus. Assume the gold nucleus remains stationary. How close does the alpha particle get to the gold nucleus before turning around? (the distance of closest approach ) 2.74 x 10-14 m 15