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 2. Electrons in - - - - - - - - - - - - - - + - + - - + + - + + + + - - - - 1. Bring negatively charged strip near neutral ES 2. Electrons in plate are repelled to bottom of ES; stick is repelled from stand 3. Touch strip to ES 4. Take strip away; ES is negatively charged - - - - - - - + + + + - - - - - - - - - - - - Electrons transfer from strip to ES

(induces charge movement; acquires charge 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 e- + - Ground connection is broken; ball is now positive + + + + Take strip away; left with positively charged ball

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

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 Coulomb’s Law: q1 = first charge q2 = second charge r = distance Mathematically, k q1 q2 F = r2 k = 9 x 109 N m2/C2 Unit of charge: 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. k q1 q2 F = 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 F F = force q = charge E = q

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 F = 12 N F 12 N E = = q 4.0 x 10-6 C E = 3.0 x 106 N/C

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: + +

the work done per unit charge as a charge is Potential Difference: the work done per unit charge as a charge is moved between two locations in an electric field; denoted by V W Potential Difference V = 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. W 24 J = = 12 J/C V = q 2.0 C = 12 volts = 12 V Alessandro Volta Definition: 1 volt = 1 V = 1 J/C ( 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 W W = q V V = q = ( 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 me = 9.11 x 10-31 kg 2 KE 2 ( 1.92 x 10-17 J ) v = = m 9.11 x 10-31 kg v = 6.50 x 106 m/s

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

Amount of stored charge depends on: (1) voltage of battery + + + + + + + + + + + + V - - - - - - - - - - - - Amount of stored charge depends on: (1) voltage of battery (2) size of capacitor Q = charge V = voltage C = capacitance Q = C V

ex. When hooked to a 12-V battery, 6.0 mC of charge is stored on a capacitor. Find the capacitance. Q = C V Q 6.0 x 10-6 C C = = V 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: (1) voltage of battery (2) distance between plates Electric Field in a Capacitor V E = 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. V 12 V E = = d 0.0015 m E = 8000 V/m