AP Electrostatics The force between two isolated charges is governed by Coulomb’s Law: F e = k q 1 q 2 r2r2 q 1 and q 2 are charges r = distance k = 9 x 10 9 Nm 2 /C 2 More generally: F e = q 1 q 2 r2r2 4π εo4π εo 1 ε o = permittivity of free space = 8.85 x C 2 /Nm 2

1. Use Coulomb’s Law to determine the force between a charge of +3.5  C and –2.0  C when they are separated by a distance of 0.50 cm. q 1 = +3.5 μC= +3.5 x C q 2 = μC= x C r = 0.50 cm = m F e = q 1 q 2 r2r2 4π εo4π εo 1ε o = 8.85 x C 2 /Nm 2 = 4π (8.85 x C 2 /Nm 2 ) ( m ) 2 1 (+3.5 x C)(-2.0 x C) = 2520 N

Electric Fields Consider a point charge +Q : +Q Charge will generate an electric field Field can be “mapped” by the trajectory of a small charge +q placed in the region around +Q +q Strength of the field is determined by the amount of force on the charge Field lines begin on positive charges and end on negative charges Field lines never intersect

Electric field intensity E = F q 2. A charge of 4.5 mC feels a force of 6.5 N when in an electric field. Find the electric field intensity. E = F q = 6.5 N C E = 1444 N/C = 1400 N/C

Electric Field Around a Point Charge E = F q +Q Apply Coulomb’s Law to +q to get E = Q r2r2 4π εo4π εo 1 Electric fields are vectors; have a direction

3. Consider a sphere of charge +4.0 μC. Consider the sphere as a point charge. (a) Find the electric field intensity 0.60 m away from the sphere. E = Q r2r2 4π εo4π εo 1 = k Q r2r2 = ( 9 x 10 9 Nm 2 /C 2 )( 4.0 x C ) ( 0.60 m ) 2 E = 1.0 x 10 5 N/C, away from sphere

Electric Potential Defined as: the work done per unit charge as the charge is moved in an electric field +Q +q V = W q V = - Q r4π εo4π εo 1 Around a Point Charge: negative, because field pushes charge away, increasing its KE, so decreasing its PE V = - k Q r OR :

3. Consider a sphere of charge +4.0 μC. Consider the sphere as a point charge. (b) What is the electric potential 0.60 m away from the sphere? V = - k Q r = - ( 9 x 10 9 Nm 2 /C 2 )( 4.0 x C ) ( 0.60 m ) V = Nm/C = J/C = V Electric potential is a scalar quantity: no direction, just a magnitude

3. Consider a sphere of charge +4.0 μC. Consider the sphere as a point charge. (c) If an electron were placed 0.60 m away from the sphere, what would be its electric potential energy? Electric potential at that location = V = J/C To get energy (in joules), multiply by the charge So PE e = q V= ( x C )( J/C ) PE e = 9.6 x J

U e = - q 1 q 2 r4π εo4π εo 1 = - k q 1 q 2 r Electric Potential Energy In general, the PE of one charge q 1 due to a second charge q 2 : PE e = q 1 V 2 V 2 = - q2q2 r4π εo4π εo 1 On AP Test Equation Bank, U represents potential energy

4. Consider the system below: (a) Find the net electric force on charge in the middle μC μC+ 1.0 μC 8.0 cm4.0 cm F 13 = k q 1 q 3 r2r2 = ( m ) 2 (9 x 10 9 Nm 2 /C 2 )(+4.0 x C)(-1.0 x C) = Nto the left N F 23 = k q 2 q 3 r2r2 = ( m ) 2 (9 x 10 9 Nm 2 /C 2 )(-1.0 x C)(+1.0 x C) = Nto the right N F net = ( N ) + ( N )= F net = 0 q1q1 q3q3 q2q2

4. Consider the system below: (b) Find the net electric field at pt. A μC μC 8.0 cm4.0 cm E 1 = k q 1 r2r2 = ( m ) 2 (9 x 10 9 Nm 2 /C 2 )(+4.0 x C) = x 10 6 N/Cto the right E2E2 E 2 = k q 2 r2r2 = ( m ) 2 (9 x 10 9 Nm 2 /C 2 )(+1.0 x C) = x 10 6 N/Cto the left E1E1 E net = ( E6 N ) + ( E6 N )= E net = 0 A q1q1 q2q2

4. Consider the system below: (c) Find the electric potential at pt. A μC μC 8.0 cm4.0 cm A V 1 = - k q 1 r = m (9 x 10 9 Nm 2 /C 2 )(+4.0 x C) = V V 2 = - k q 2 r = m (9 x 10 9 Nm 2 /C 2 )(+1.0 x C) = V q1q1 q2q2

Because electric potential is a scalar quantity, just add the individual potentials to find the net potential: V net = V 1 + V 2 = ( V ) + ( V ) V net = V 4. Consider the system below: (c) Find the electric potential at pt. A μC μC 8.0 cm4.0 cm A V 1 = V V 2 = V q1q1 q2q2

5. (a) Determine the net electric field at pt. A. A C C 36 cm 15 cm E1E1 E 1 = k Q 1 Electric field at A due to the +2.0-C charge: = ( 9 x 10 9 Nm 2 /C 2 )( 2.0 C ) ( 0.15 m ) 2 E 1 = 8.0 x N/C, in positive y-direction r12r12

5. (a) Determine the net electric field at pt. A. A C C 36 cm 15 cm E1E1 E 2 = k Q 2 r22r22 Electric field at A due to the -3.0-C charge: = ( 9 x 10 9 Nm 2 /C 2 )( C ) ( 0.39 m ) 2 E 2 = x N/C E2E2 r r 2 = (15) 2 + (36) 2 r 2 = 1521 r = 39 Net electric field will be the vector sum of the individual fields Add the vectors by the component method or better, E 2 = x N/C, towards the charge in the direction shown

A C C 36 cm 15 cm E1E1 E2E2 E 2x E 2y θ θ Draw x- and y-components of E 2 ; need angle θ ; from geometry, we know small angle in other triangle = θ From SOH-CAH-TOA:tan θ = = θ = tan -1 ( )= 22.6 o E 1 = 8.0 x N/CE 2 = x N/C Then cos 22.6 = E 2x x E 2x = 1.64 x sin 22.6 = E 2y x E 2y = 6.82 x and

A C C 36 cm 15 cm E1E1 E2E2 E 2x E 2y θ θ E 1y = 8.0 x N/C E 2x = 1.64 x N/C E 2y = x N/C ( downward ) E 1x = 0 x- and y-components of resultant: R x = E 1x + E 2x = 0 + ( 1.64 x ) = 1.64 x R y = E 1y + E 2y = ( 8.0 x ) + ( x ) = x 10 11

A C C 36 cm 15 cm E1E1 E2E2 E 2x E 2y θ R y = x R x = 1.64 x RyRy RxRx R R 2 = ( 1.64 x ) 2 + ( x ) 2 R 2 = 5.62 x R = 7.5 x RyRy tan θ = x = 4.46 θ = tan -1 ( 4.46 )= 77 o 1.64 x So R = 7.5 x N/C at 77 o to the horizontal

5. (b) Determine the electric potential at pt. A. A C C 36 cm 15 cm V 1 = - k Q 1 r1r1 Potential at A due to the +2.0-C charge: = - ( 9 x 10 9 Nm 2 /C 2 )( 2.0 C ) ( 0.15 m ) V 1 = x V V 2 = - k Q 2 r2r2 Potential at A due to the C charge: = - ( 9 x 10 9 Nm 2 /C 2 )( C ) ( 0.39 m ) V 2 = x V 39 cm

5. (b) Determine the electric potential at pt. A. A C C 36 cm 15 cm V 1 = x V Because electric potential is a scalar quantity, just add the individual potentials to find the net potential: V 2 = x V 39 cm V net = V 1 + V 2 = ( x V ) + ( x V ) V net = 5.1 x V

6. In Millikan’s oil-drop experiment, a drop of oil was suspended between two parallel plates oriented horizontally. If the drop has a weight of 2.6 x N and it has a charge of –3q, where q is the charge on an electron, determine the strength of the electric field between the plates FeFe Wt. = q E q E = Wt. q E = Wt. q = 2.6 x N -3( x C ) E = 5.4 x N/C, downward -3q

Other general points: But first, a definition: an equipotential line is a line that connects places of equal potential in an electric field ( Gravitational analog: the surface of a table is a gravitational equipotential surface, since every place on the table is the same height off the floor, and so has the same gravitational PE ) Electric field lines are always perpendicular to equipotential lines ( Again, relate to gravity: to map an electric field, we placed a charge in the field and released it; to map a gravitational field, place an object in the field and release it [hold your pencil above the table and release it]; the pencil will hit the table along a line perpendicular to the table; so the gravitational field line [the path of the pencil] is perpendicular to the equipotential surface [the table])

All static charge on a conductor resides on its surface _ _ _ _ _ _ _ _ If charge is placed on a conductor, it can move around within the conductor Like charges repel, so they will get as far from each other as possible (on the surface) The surface of a conductor is an equipotential surface If charge is placed on a conductor, it can move around on its surface ( can you think of a gravitational analog?) ; if some of the charge is at a higher potential, it will move until it is at the same potential as all other charges

No electric field can exist on the inside of a conductor On non-spherical conductors, charge congregates at points of high curvature; the higher the curvature, the higher the charge density On spherical conductors, charge distributes itself evenly (density of charge is equal everywhere)

A charge +Q is placed inside a spherical conducting shell; +Q what would the resulting electric field look like? +Q charge will induce a charge of –Q on the inner surface Q charge will then be on the outside surface

+Q Resulting electric field:

Capacitors Devices that store electric charge Consist of two metal plates that are parallel to each other and separated by a small distance V When connected to a battery, charge will flow out of the battery and onto the plates Will be stored there indefinitely

Two metal plates of area A are separated by a distance d A d The capacitance of the capacitor (related to the amount of charge it can hold) depends on: - the area Acapacitance increases directly with area - the separation distance d the closer the plates are, the greater the capacitance - the material between the plates the presence of a dielectric increases the capacitance

A d Summarized by: C = K ε o A d C = capacitance A = area of plates d = separation distance ε o = 8.85 x C 2 /Nm 2 K = dielectric constant

ex. Two parallel plates of cross-sectional area 5.0 cm 2 are separated by a distance of 0.50 cm. Determine the capacitance. C = K ε o A d A = 5.0 cm 2 = m 2 d = m Dielectric assumed to be air : K = 1 = ( 1 )( 8.85 x C 2 /Nm 2 )( m 2 ) m C = 8.85 x C 2 /Nm = 8.85 x farads = 8.85 x F 1 C 2 / Nm Units fun: = 1 C 2 / J = 1 C / J/C = 1 C / V Definition: 1 farad = 1 F = 1 C/V

V When connected to a battery, a certain amount of charge is stored on the capacitor The amount of charge stored on the capacitor depends on: - the capacitance C - the voltage of the battery Q = C V Q = charge C = capacitance V = voltage

ex. If the capacitor in the previous example is placed across a 6-V battery, find the charge placed on the plates. C = 8.85 x FV = 6.0 VQ = ? Q = C V = ( 8.85 x F )( 6.0 V ) = ( 8.85 x C/V )( 6.0 V ) Q = 5.3 x C

Energy Stored in a Charged Capacitor Q = C V The energy stored in a charged capacitor can be found by the expression E c = ½ Q V = ½ C V 2 ex. Determine the energy stored in the capacitor in the previous example. C = 8.85 x FV = 6.0 V E c = ½ C V 2 = ½ ( 8.85 x F )( 6.0 V ) 2 E c = 1.6 x J Units check:F V 2 = C/V V 2 = C V = C J/C = J

Electric Field in a Charged Capacitor V Electric field lines begin on positive charges and end on negative charges So field lines begin on positive plate and end on negative plate Away from the edges, field lines are parallel ; at edges field lines arc from positive plate to negative plate Within the capacitor (away from the edges), field is uniform throughout the region; has the same intensity near the positive plate, near the negative plate, or right in the middle

Electric Field in a Charged Capacitor V d The strength of the electric field (electric field intensity) depends on: - voltage V of the battery (higher voltage, stronger field) - separation distance d (if plates are closer together, field is more intense) E = V d

ex. Determine the electric field intensity inside the capacitor in the previous examples. V = 6.0 V d = 0.50 cm = m E = V d = 6.0 V m E = 1200 V/m