AP Electrostatics The force between two isolated charges is governed by Coulomb’s Law: q1 and q2 are charges k q1 q2 Fe = r = distance r2 k = 9 x 109 Nm2/C2 More generally: εo = permittivity of free space 1 q1 q2 Fe = 4π εo r2 = 8.85 x 10-12 C2/Nm2

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. q1 = +3.5 μC = +3.5 x 10-6 C r = 0.50 cm = 0.0050 m q2 = - 2.0 μC = - 2.0 x 10-6 C 1 q1 q2 εo = 8.85 x 10-12 C2/Nm2 Fe = 4π εo r2 1 (+3.5 x 10-6C)(-2.0 x 10-6C) = 4π (8.85 x 10-12 C2/Nm2) ( 0.0050 m )2 = 2520 N

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

Electric field intensity F E = 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. F 6.5 N E = = q 0.0045 C E = 1444 N/C = 1400 N/C

Electric Field Around a Point Charge Apply Coulomb’s Law to F +q E = q +Q to get 1 Q E = 4π εo r2 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. 1 Q k Q E = = 4π εo r2 r2 ( 9 x 109 Nm2/C2 )( 4.0 x 10-6 C ) = ( 0.60 m )2 E = 1.0 x 105 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 W V = +Q q Around a Point Charge: negative, because field pushes charge away, increasing its KE, so decreasing its PE 1 Q V = - 4π εo r - k Q V = OR : r

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? - k Q V = r - ( 9 x 109 Nm2/C2 )( 4.0 x 10-6 C ) = ( 0.60 m ) V = - 60 000 Nm/C Electric potential is a scalar quantity: no direction, just a magnitude = - 60 000 J/C = - 60 000 V

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 = - 60 000 V = - 60 000 J/C To get energy (in joules), multiply by the charge So PEe = q V = ( - 1.6021x 10-19 C )( - 60 000 J/C ) PEe = 9.6 x 10-15 J

In general, the PE of one charge q1 due to a second charge q2 : q2 1 PEe = q1 V2 V2 = - 4π εo r On AP Test Equation Bank, U represents potential energy Electric Potential Energy 1 q1 q2 - k q1 q2 Ue = - = 4π εo r r

4. Consider the system below: (a) Find the net electric force on charge in the middle. + 4.0 μC - 1.0 μC + 1.0 μC 5.625 N 5.625 N q1 q3 q2 8.0 cm 4.0 cm k q1 q3 (9 x 109 Nm2/C2)(+4.0 x 10-6 C)(-1.0 x 10-6 C) F13 = = r2 ( 0.080 m )2 = 5.625 N to the left k q2 q3 (9 x 109 Nm2/C2)(-1.0 x 10-6 C)(+1.0 x 10-6 C) F23 = = r2 ( 0.040 m )2 = 5.625 N to the right Fnet = ( + 5.625 N ) + ( - 5.625 N ) = Fnet = 0

4. Consider the system below: (b) Find the net electric field at pt. A. + 4.0 μC + 1.0 μC E2 E1 q1 A q2 8.0 cm 4.0 cm k q1 (9 x 109 Nm2/C2)(+4.0 x 10-6 C) E1 = = r2 ( 0.080 m )2 = 5.625 x 106 N/C to the right k q2 (9 x 109 Nm2/C2)(+1.0 x 10-6 C) E2 = = r2 ( 0.040 m )2 = 5.625 x 106 N/C to the left Enet = ( + 5.625E6 N ) + ( - 5.625E6 N ) = Enet = 0

4. Consider the system below: (c) Find the electric potential at pt. A. + 4.0 μC + 1.0 μC q1 A q2 8.0 cm 4.0 cm k q1 (9 x 109 Nm2/C2)(+4.0 x 10-6 C) V1 = - = - r 0.080 m = - 450 000 V k q2 (9 x 109 Nm2/C2)(+1.0 x 10-6 C) V2 = - = - r 0.040 m = - 225 000 V

4. Consider the system below: (c) Find the electric potential at pt. A. + 4.0 μC + 1.0 μC q1 A q2 8.0 cm 4.0 cm V1 = - 450 000 V V2 = - 225 000 V Because electric potential is a scalar quantity, just add the individual potentials to find the net potential: Vnet = V1 + V2 = ( - 450 000 V ) + ( - 225 000 V ) Vnet = - 675 000 V

5. (a) Determine the net electric field at pt. A. Electric field at A due to the +2.0-C charge: E1 k Q1 A E1 = r12 15 cm ( 9 x 109 Nm2/C2 )( 2.0 C ) 36 cm = ( 0.15 m )2 + 2.0 C - 3.0 C E1 = 8.0 x 1011 N/C , in positive y-direction

5. (a) Determine the net electric field at pt. A. Electric field at A due to the -3.0-C charge: r2 = (15)2 + (36)2 E1 r2 = 1521 r = 39 k Q2 A E2 = E2 r22 r 15 cm ( 9 x 109 Nm2/C2 )( - 3.0 C ) 36 cm = ( 0.39 m )2 + 2.0 C - 3.0 C E2 = - 1.775 x 1011 N/C or better, E2 = 1.775 x 1011 N/C , towards the charge in the direction shown Net electric field will be the vector sum of the individual fields Add the vectors by the component method

Draw x- and y-components of E2 E1 = 8.0 x 1011 N/C E2 = 1.775 x 1011 N/C Draw x- and y-components of E2 ; from geometry, we know small angle in other triangle = θ ; need angle θ E1 15 From SOH-CAH-TOA: tan θ = = 0.417 36 E2x A θ θ = tan -1 ( 0.417 ) = 22.6o E2y E2 15 cm 36 cm θ + 2.0 C - 3.0 C Then and E2x E2y cos 22.6 = sin 22.6 = 1.775 x 1011 1.775 x 1011 E2x = 1.64 x 1011 E2y = 6.82 x 1010

x- and y-components of resultant: E1x = 0 E1 E1y = 8.0 x 1011 N/C E2x E2x = 1.64 x 1011 N/C A θ E2y = - 6.82 x 1010 N/C ( downward ) E2y E2 15 cm 36 cm θ + 2.0 C - 3.0 C x- and y-components of resultant: Rx = E1x + E2x = 0 + ( 1.64 x 1011 ) = 1.64 x 1011 Ry = E1y + E2y = ( 8.0 x 1011 ) + ( - 6.82 x 1010 ) = 7.318 x 1011

So R = 7.5 x 1011 N/C at 77o to the horizontal Rx = 1.64 x 1011 Ry = 7.318 x 1011 E1 R E2x Ry Ry A E2y E2 15 cm θ 36 cm Rx + 2.0 C - 3.0 C 7.318 x 1011 R 2 = ( 1.64 x 1011 )2 + ( 7.318 x 1011 )2 tan θ = = 4.46 1.64 x 1011 R 2 = 5.62 x 1023 θ = tan -1 ( 4.46 ) = 77o R = 7.5 x 1011 So R = 7.5 x 1011 N/C at 77o to the horizontal

5. (b) Determine the electric potential at pt. A. Potential at A due to the +2.0-C charge: A 39 cm 15 cm k Q1 V1 = - 36 cm r1 + 2.0 C - 3.0 C ( 9 x 109 Nm2/C2 )( 2.0 C ) = - ( 0.15 m ) V1 = - 1.2 x 1011 V Potential at A due to the - 3.0-C charge: k Q2 ( 9 x 109 Nm2/C2 )( - 3.0 C ) V2 = - = - r2 ( 0.39 m ) V2 = + 6.92 x 1010 V

5. (b) Determine the electric potential at pt. A. V1 = - 1.2 x 1011 V 39 cm 15 cm V2 = + 6.92 x 1010 V 36 cm + 2.0 C - 3.0 C Because electric potential is a scalar quantity, just add the individual potentials to find the net potential: Vnet = V1 + V2 = ( - 1.2 x 1011 V ) + ( + 6.92 x 1010 V ) Vnet = - 5.1 x 1010 V

But first, a definition: an equipotential line is a line that 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 ; if some of the charge is at a higher potential, it will move until it is at the same potential as all other charges ( can you think of a gravitational analog?)

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

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

Resulting electric field: + + + +Q + + + + +

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 The capacitance of the capacitor (related to the amount of charge it can hold) depends on: - the area A capacitance 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 = capacitance K εo A A = area of plates C = d d = separation distance εo = 8.85 x 10-12 C2/Nm2 K = dielectric constant

a distance of 0.50 cm. Determine the capacitance. ex. Two parallel plates of cross-sectional area 5.0 cm2 are separated by a distance of 0.50 cm. Determine the capacitance. A = 5.0 cm2 = 0.00050 m2 d = 0.0050 m Dielectric assumed to be air : K = 1 K εo A ( 1 )( 8.85 x 10-12 C2/Nm2 )( 0.00050 m2 ) C = = d 0.0050 m Units fun: C = 8.85 x 10-13 C2/Nm 1 C2 / Nm = 1 C2 / J = 8.85 x 10-13 farads = 1 C / J/C = 8.85 x 10-13 F = 1 C / V Definition: 1 farad = 1 F = 1 C/V

When connected to a battery, a certain amount of charge is stored on the capacitor + + + + + + + V The amount of charge stored on the capacitor depends on: - the capacitance C - the voltage of the battery Q = charge Q = C V 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 10-13 F V = 6.0 V Q = ? Q = C V = ( 8.85 x 10-13 F )( 6.0 V ) = ( 8.85 x 10-13 C/V )( 6.0 V ) Q = 5.3 x 10-12 C

Energy Stored in a Charged Capacitor The energy stored in a charged capacitor can be found by the expression Ec = ½ Q V = ½ C V2 Q = C V ex. Determine the energy stored in the capacitor in the previous example. C = 8.85 x 10-13 F V = 6.0 V Ec = ½ C V2 = ½ ( 8.85 x 10-13 F )( 6.0 V )2 Ec = 1.6 x 10-11 J Units check: F V2 = C/V V2 = 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 + + + + + + + + + d V 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) V E = d

ex. Determine the electric field intensity inside the capacitor in the previous examples. V = 6.0 V d = 0.50 cm = 0.0050 m V E = d 6.0 V = 0.0050 m E = 1200 V/m

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 10-6 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. q E = Wt. q q + + + + + + + + + Fe = q E Wt. E = -3q q Wt. 2.6 x 10-6 N = -3( 1.6021 x 10-19 C ) E = 5.4 x 1012 N/C , downward

Capacitors in Series and Parallel (1) Q = Q1 = Q2 = Q3 C1 - + (2) V = V1 + V2 + V3 - + - Q + - + + + + + Q1 = C1 V1 C2 V - - - - Q2 = C2 V2 - + - + Q3 = C3 V3 Q = Ceq V - + - + C3 Q Q1 V1 = V = C1 Ceq Q Q2 Q3 V2 = V3 = Ceq V C2 C3

Series: (1) Q = Q1 = Q2 = Q3 C1 (2) V = V1 + V2 + V3 Q Q V = C2 V Ceq - + (2) V = V1 + V2 + V3 - + - Q + - Q + + + + + V = C2 V Ceq - - - - - + - + Q1 Q2 Q - + V = + + - + C1 C2 C3 C3 Q Q Q Q + = + = C1 C2 C3 Ceq Q 1 1 1 1 Ceq + + V = Ceq C1 C2 C3

Parallel: Q Q V C1 C2 C3 V Ceq (1) V = V1 = V2 = V3 Q1 = C1 V1 + + + + + + + + + V C1 C2 C3 V Ceq - - - - - - - - - (1) V = V1 = V2 = V3 Q1 = C1 V1 (1) Q = Q1 + Q2 + Q3 Q2 = C2 V2 Q = Ceq V Q3 = C3 V3 Q = C1V1 + C2V2 + C3V3 = C1V + C2V + C3V = Ceq V Ceq = C1 + C2 + C3

Kirchoff’s Rules for Circuit Analysis (1) Current into a junction = Current out of the junction (2) The sum of the changes in potential around any closed loop of a circuit = 0 ex. Find the currents I1 , I2 , and I3 . 9.0 V I1 A junction is a place in the circuit where the current splits, or comes together. (see highlighted places) RB RC I2 I3 100 Ω 100 Ω 15 V RA 100 Ω There are three unknowns ( I1 , I2 , and I3 ); we will need 3 independent equations to find them; we will use Kirchoff’s Rules to derive the 3 equations, and then solve them for I1 , I2 , and I3

(1) Current into a junction = Current out of the junction Kirchoff’s Rules: (1) Current into a junction = Current out of the junction (2) The sum of the changes in potential around any closed loop of a circuit = 0 9.0 V I1 I3 Notice that the 9.0-V battery is in series with RC; so the current through the battery is also I3 RB RC I2 I3 100 Ω 100 Ω 15 V RA 100 Ω Then the current into the red junction is I1 , and the current out of the junction is the sum of I2 and I3 I1 = I2 + I3

(1) Current into a junction = Current out of the junction Kirchoff’s Rules: (1) Current into a junction = Current out of the junction (2) The sum of the changes in potential around any closed loop of a circuit = 0 (1) I1 = I2 + I3 9.0 V I1 An example of a closed loop is shown; there are actually 3 loops to choose from RB RC I2 I3 100 Ω 100 Ω 15 V RA 100 Ω

problems it is useful to equate current in a circuit (1) I1 = I2 + I3 9.0 V I1 + - When analyzing these problems it is useful to equate current in a circuit with current in a river RB RC + I2 I3 100 Ω 100 Ω 15 V - RA 100 Ω If the current goes as I2 through RB, it is like the current is flowing “downhill” (river analogy), and so the potential decreases But if the current goes through RC as shown, it is like going “against the current”, or “uphill”, and so the potential increases For current through a battery: when the current flows from the positive terminal to the negative (green arrow in diagram), it decreases potential (uphill to downhill); when it goes from the negative terminal to the positive (violet arrow above), it is going “uphill”, and increases potential

Consider red loop; start at negative terminal of battery (1) I1 = I2 + I3 9.0 V I1 + - Consider red loop; start at negative terminal of battery and go around clockwise RB RC + I2 I3 100 Ω 100 Ω 15 V - RA RA is in series with the 15-V battery, so the current through RA is also I1 100 Ω I1 Potential is measured in volts; it will increase or decrease through a resistor according to Ohm’s Law ( V = I R ) Start at negative terminal: + 15 - I2 RB - I1 RA = 0 Kirchoff Rule #2 uphill through battery goes with current through RB same for the current through RA so: 15 - 100 I2 - 100 I1 = 0

Consider blue loop; start at negative terminal of battery (1) I1 = I2 + I3 9.0 V I1 + - (2) 15 - 100 I2 - 100 I1 = 0 RB RC + I2 I3 100 Ω 100 Ω Consider blue loop; start at negative terminal of battery and go around counter-clockwise 15 V - RA 100 Ω I1 Start at negative terminal: + 9 - I2 RB + I3 RC = 0 uphill through battery goes with current through RB goes against current through RC, and so increases potential so: 9 - 100 I2 + 100 I3 = 0 This is the third independent equation, along with (1) and (2) derived earlier (and listed above); the task is now to solve these equations

Note: You must use both of Kirchoff’s Rules (1) I1 = I2 + I3 Note: You must use both of Kirchoff’s Rules to get the equations; if you use only one of the Rules, the equations are not independent and won’t give correct answers (2) 15 - 100 I2 - 100 I1 = 0 (3) 9 - 100 I2 + 100 I3 = 0 Substitute (1) into (2): 15 - 100 I2 - 100 ( I2 + I3 ) = 0 Use Distributive Property 15 - 100 I2 - 100 I2 - 100 I3 = 0 Combine like terms 15 - 200 I2 - 100 I3 = 0 Add equation (3) (3) 9 - 100 I2 + 100 I3 = 0 24 - 300 I2 = 0 24 = 300 I2 I2 = 0.080 A 15 - ( 200 )( 0.080 ) - 100 I3 = 0 15 - 16 - 100 I3 = 0 - 1 - 100 I3 = 0 - 1 = 100 I3 I3 = - 0.010 A

Negative sign for current indicates that the current (1) I1 = I2 + I3 9.0 V I1 (2) 15 - 100 I2 - 100 I1 = 0 (3) 9 - 100 I2 + 100 I3 = 0 RB RC I2 I3 100 Ω 100 Ω 15 V RA I2 = 0.080 A 100 Ω I3 = - 0.010 A Negative sign for current indicates that the current flows in a direction opposite that shown in the diagram; that’s okay—don’t change it Finally, to find I1 : I1 = I2 + I3 = ( 0.080 ) + ( - 0.010 ) I1 = 0.070 A To be sure of the answers, check see next slide

(1) I1 = I2 + I3 I1 = 0.070 A (2) 15 - 100 I2 - 100 I1 = 0 I2 = 0.080 A (3) 9 - 100 I2 + 100 I3 = 0 I3 = - 0.010 A (1) I1 = I2 + I3 ? 0.070 = 0.080 + ( - 0.010 ) = 0.070 that was easy (2) 15 - 100 I2 - 100 I1 = 0 ? 15 - 100 ( 0.080 ) - 100 ( 0.070 ) = 0 ? 15 - 8 - 7 = 0 so far so good (3) 9 - 100 I2 + 100 I3 = 0 ? 9 - 100 ( 0.080 ) + 100 ( - 0.010 ) = 0 ? 9 - 8 + ( - 1 ) = 0 9 - 8 - 1 = 0 yes !!!!

ex. Find the current in each resistor. Choose currents in any direction you wish; we’ll choose them like this: R1 = 3.9 Ω R3 = 6.7 Ω I1 I3 9.0 V I2 R2 = 1.2 Ω 12 V I1 Note: the current through R4 is the same as the current through R1, since they are in series R4 = 9.8 Ω Junction Rule: (1) I1 = I2 + I3 Loop Rule: + 12 - 3.9 I1 - 1.2 I2 - 9.8 I1 = 0 red loop (2) 12 - 13.7 I1 - 1.2 I2 = 0 + 12 - 3.9 I1 - 6.7 I3 - 9 - 9.8 I1 = 0 green loop (3) 3 - 13.7 I1 - 6.7 I3 = 0

(1) I1 = I2 + I3 (2) 12 - 13.7 I1 - 1.2 I2 = 0 (3) 3 - 13.7 I1 - 6.7 I3 = 0 Solve these equations by previous methods for I1 , I2 , and I3 OR , use matrix arithmetic and let your TI-84 calculator solve them Here’s how: Manipulate the equations so that all of the unknowns are on one side and the constants are on the other side; put the unknowns in the same order I1 = I2 + I3 I1 - I2 - I3 = 0 12 - 13.7 I1 - 1.2 I2 = 0 13.7 I1 + 1.2 I2 = 12 3 - 13.7 I1 - 6.7 I3 = 0 13.7 I1 + 6.7 I3 = 3

I1 - I2 - I3 = 0 1 I1 - 1 I2 - 1 I3 = 0 13.7 I1 + 1.2 I2 = 12 13.7 I1 + 1.2 I2 + 0 I3 = 12 13.7 I1 + 6.7 I3 = 3 13.7 I1 + 0 I2 + 6.7 I3 = 3 re-write with coefficients for each unknown make into a matrix 1. Find Matrix in your calculator and enter the values in a 3 x 4 matrix 1 - 1 - 1 0 13.7 1.2 0 12 2. Find the option “rref” and do it on your matrix ex. rref ( [A] ) if A is the name of the matrix 13.7 0 6.7 3 3. Answers will be in the last column 1 0 0 0.722 I1 = 0.722 A 0 1 0 1.75 I2 = 1.75 A let’s check 0 0 1 - 1.03 I3 = - 1.03 A

(1) I1 - I2 - I3 = 0 I1 = 0.722 A I2 = 1.75 A (2) 13.7 I1 + 1.2 I2 = 12 I3 = - 1.03 A (3) 13.7 I1 + 6.7 I3 = 3 (1) I1 - I2 - I3 = 0 ? (0.722) - (1.75) - (-1.03) = (0.722) - (1.75) + (1.03) = 0 I expected that (2) 13.7 I1 + 1.2 I2 = 12 ? 13.7 ( 0.722 ) + 1.2 ( 1.75 ) = 12 ? 9.89 + 2.1 = 12 oh my goodness….. (3) 13.7 I1 + 6.7 I3 = 3 ? 13.7 ( 0.722 ) + 6.7 ( - 1.03 ) = 3 ? 9.89 + ( - 6.9 ) = 9.89 - 6.9 = 3 sweet

Immediately after switch is closed (t = 0) what is the current RC Circuits R3 = 3 Ω I3 4 µF R2 = 12 Ω I2 18 V I1 R1 = 6 Ω Immediately after switch is closed (t = 0) what is the current through each resistor?

Immediately after switch is closed (t = 0) what is the current RC Circuits R3 = 3 Ω I3 4 µF R2 = 12 Ω I2 18 V I1 R1 = 6 Ω Immediately after switch is closed (t = 0) what is the current through each resistor? Initially, capacitor acts as a short circuit, drawing all current from R2 I2 = 0

I2 = 0 I1 = I3 = 2 A V 18 V I = = = 2 A Req 9 Ω R3 = 3 Ω I3 4 µF I2

After the switch has been closed a long time (t = ∞) what is the 18 V I1 R1 = 6 Ω After the switch has been closed a long time (t = ∞) what is the current through each resistor? After the capacitor is charged, it acts like an open circuit

I1 = I2 = I3 = 0.86 A V 18 V I = = = 0.86 A Req 21 Ω R3 = 3 Ω I3 4 µF

After the switch has been closed a long time (t = ∞) what is the Vc 18 V I1 R1 = 6 Ω After the switch has been closed a long time (t = ∞) what is the charge on the capacitor? Q = C VC

I1 = I2 = I3 = 0.86 A VC = I2 R2 = ( 0.86 A )( 12 Ω ) = 10.3 V 4 µF R2 = 12 Ω I2 Vc 18 V I1 R1 = 6 Ω R3 = 3 Ω VC = I2 R2 = ( 0.86 A )( 12 Ω ) R2 = 12 Ω I2 Vc 18 V = 10.3 V I1 R1 = 6 Ω

After the switch has been closed a long time (t = ∞) what is the Vc 18 V I1 R1 = 6 Ω After the switch has been closed a long time (t = ∞) what is the charge on the capacitor? Q = C VC = ( 4 µF )( 10.3 V ) = Q = 41 µC