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Electrostatics •Objects become charged due to the movement of electrons •Protons are locked in the nucleus and cannot move conductor: a material on which electrons can move easily example: metals • Metals conduct because outermost electrons are loosely held; electrons are “bees” and the atoms are “beehives” insulator: a material through which electrons cannot move • Insulators’ electrons are locked in the atom; can’t get out examples: glass plastic Results from Electrostatics Lab A negatively charged strip is brought near a neutral ball + - + Electrons are repelled to far side; protons are closer, so it attracts + + - After touching, electrons move from strip to ball Strip and ball are no both negative, so they repel each other + + - - - • Charged objects attract neutral objects •After neutral objects touch charged objects, charge transfer occurs; acquire like charge 3 Ways to Charge Objects • Friction ( plastic rod in lab, walking across carpet ) • Conduction 3 Ways to Charge Objects • Friction ( plastic rod in lab, walking across carpet ) • Conduction 3 Ways to Charge Objects • Friction ( plastic rod in lab, walking across carpet ) • Conduction (contact; acquires same charge as charging object) electroscope: plate stick stand - - - - - - - - - - - - - 1. Bring negatively charged strip near neutral ES - - - - - - - - - - - - + - + - + + - 3. Touch strip to ES - - - - - 4. Take strip away; ES is negatively charged - - - - - - - - - - - - - ++++ Electrons transfer from strip to ES 2. Electrons in plate are repelled to bottom of ES; stick is repelled from stand + + + + - - - - - - • Induction (induces charge movement; acquires charge opposite that of charging object) A negatively charged strip is brought near, but not touching, a neutral ball + - + Electrons are repelled to far side + + - Connection to ground is made; electrons are repelled to ground + + - Ground connection is broken; ball is now positive ++ ++ Take strip away; left with positively charged ball ++ ++ e- 1. Bring a negatively charged strip near two connected metal spheres that are neutral and insulated from the ground - - - - - - - - - - - - - A B 2. While the strip is held near (but not touching), separate the spheres - - - - - - - - - - - - - A B 3. Take the strip away; what is the charge on each sphere? A B Coulomb’s Law: The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them Mathematically, F = Unit of charge: k q1 q2 r2 q1 q2 r k = = = = first charge second charge distance 9 x 109 N m2/C2 coulomb Charge of an electron: 1.6021 x 10-19 C Charles Coulomb ( 1736 - 1806 ) ex. Find the force between charges of +1.0 C and -2.0 C located 0.50 m apart in air. F = = k q1 q2 r2 ( 9 x 109 Nm2/C2 )( +1.0 C )( -2.0 C ) ( 0.50 m)2 F = - 7.2 x 1010 N negative sign indicates attraction Electric Fields An electric field is said to exist in a region of space if an electric charge in the region is subject to an electric force The amount of force on the charge is a measure of electric field intensity Electric Field Intensity E= F q F = force q = charge ex. A charge of 4.0 µC experiences a force of 12 N when placed in an electric field. Find the electric field intensity. q = 4.0 μC = 4.0 x 10-6 C E= F q = 12 N 4.0 x 10-6 C E = 3.0 x 106 N/C F = 12 N Electric Field Lines To map an electric field, draw the trajectory of a very small positive test charge Field around a positive charge: + q+ Electric Field Lines To map an electric field, draw the trajectory of a very small positive test charge Field around a positive charge: + Field around a negative charge: _ Field around unlike charges: + _ •Field lines begin on positive charges and end on negative charges •Field lines never intersect Field around like charges: + + Potential Difference: the work done per unit charge as a charge is moved between two locations in an electric field; denoted by V Potential Difference V= W q ex. 24 J of work is done on a 2.0-C charge to move it in an electric field. Find the potential difference. V= W q = 24 J 2.0 C = 12 J/C = 12 volts = 12 V Definition: 1 volt = 1 V = 1 J/C Alessandro Volta ( 1745 - 1827 ) ex. An electron is accelerated through a potential difference of 120 V. (a) What energy does the electron acquire? By Work-Energy Theorem, Energy acquired = Work done on electron V= W q W = qV = ( 1.6021 x 10-19 C )( 120 V ) Work done = Energy acquired = 1.92 x 10-17 J (b) What speed does it acquire? Energy = 1.92 x 10-17 J KE = 1/2 mv2 v = 2 KE me = 9.11 x 10-31 kg = m v = 6.50 x 106 m/s 2 ( 1.92 x 10-17 J ) 9.11 x 10-31 kg Distribution of Charge - All static charge on a conductor resides on its surface - No electric field can exist on the inside of a conductor - There is no potential difference between any two points on a conducting surface -If charge is put on a spherical conducting surface, it will distribute itself evenly on the surface - If charge is placed on a non-spherical conductor, _ _ _ _ _ __ __ it will congregate at points of high curvature - If charge is placed on a non-spherical conductor, _ _ _ _ _ _ ___ _ ____ it will congregate at points of high curvature the higher the curvature, the greater the density of charge St. Elmo’s Fire Parallel-plate capacitors A device that will store electric charge ; made of two parallel metal plates separated by a small distance + + + Battery + + + + + + + + - - + - - - - - - - - - When connected to a battery, charge will flow to the plates and stay there indefinitely C V + + + + + + - - + + + + + + - - - Amount of stored charge depends on: - - - - - - (1) voltage of battery (2) size of capacitor Q=CV - Q = charge V = voltage C = capacitance ex. When hooked to a 12-V battery, 6.0 mC of charge is stored on a capacitor. Find the capacitance. Q=CV C= Q V = 6.0 x 10-6 C 12 V C = 5.0 x 10-7 C/V = 5.0 x 10-7 farads = 5.0 x 10-7 F Definition: 1 C/V = 1 farad = 1 F Michael Faraday ( 1791 – 1867 ) Electric Field in a Parallel-plate Capacitor V + + + + + + + + + - - - + + + - - - - - - - - Electric field has the same value anywhere within the plates (away from the edges) Easy to control; used to deflect charges Electric Field in a Parallel-plate Capacitor + + + V + + + + + + + + + d - - - Strength of field in capacitor depends on: Electric Field in a Capacitor E= - - - - - - - - (1) voltage of battery (2) distance between plates V d ex. A 250 mF capacitor, with 1.5 mm separation between the plates, is hooked to a 12-V battery. Find the electric field intensity between the plates. E = V d = E = 8000 V/m 12 V 0.0015 m