Wednesday 2 Feb:

 

 

Quiz: Capacitors on Thursday 3 Feb

Equations: Ch 15-18 on Monday 7th

Test Ch 15-18 on Thurs/Fri 10/11 Feb

 

Go over Elect Quiz #3 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PLATE AREA: greater plate area gives greater capacitance; less plate area gives less capacitance.

Explanation: Larger plate area results in more field flux (charge collected on the plates) for a given field force (voltage across the plates).

 

PLATE SPACING: Further plate spacing gives less capacitance; closer plate spacing gives greater capacitance.

Explanation: Closer spacing results in a greater field force (voltage across the capacitor divided by the distance between the plates), which results in a greater field flux (charge collected on the plates) for any given voltage applied across the plates.

 

DIELECTRIC MATERIAL: Greater permittivity of the dielectric gives greater capacitance; less permittivity of the dielectric gives less capacitance.

Explanation:  Materials with a greater permittivity allow for more field flux (offer less opposition), and thus a greater collected charge, for any given amount of field force (applied voltage).

1) Equation for relationship between Electrical Field, Force & Charge

2) Equation for relationship between Electrical Field, Volts & distance

3) Equations for Electrical  Potential

4) Equations for Potential (storied) Electrical  Energy

5) Equation for Work in Electrical Energy

 

 

 

 

 

            1) E = F/q

 

            2)  E = V/d   (electric potential / dist (m))

 

            3) V = kq/r 

 

           or   E d   (electric field)(dist)

 

           or V = IR   (Ohm's Law)

 

          or V = Q/C   (Charge) / (Capacitance)

 

4) U = 1/2QV  = 1/2(CV2)

 

    5) W = DVq = DU = D1/2QV = D1/2(CV2)

 

 

 

 

Capacitance (effective size of a capacitor):

C = Q/V     Capacitance = (charge/volts)

Q = CV

C = EoA/d

 

Charge: in series, the equivalent charge is the same for each capacitor. But with capacitors in parallel you have to add the individual charges to get the total charge.

 

How to increase capacitance (three things):

1.        Increase the  plate area (A)

--> more charge can be stored, more spacing.

 

2.       Decrease the distance between plates (d)

 --> stronger force of attraction (opposite plate charges) the closer the plates are (charges can bunch closer together)

3.       Use better dielectric:

(Change Eo to higher constant --> prohibits movement)

è    holds more charges because material won't allow the charges cross over from plate to plate.

 

Series Capacitors (Equivalent C):  Cequiv = (C1 x C2)/ (C1)+(C2)                            

Parallel Capacitors (Equivalent C):   Cequiv = (C1) + (C2) …

 

 

 

 

 

 

 

 

The following questions will be a quiz on this Thursday

 

 

 

 

 

 

 

 

 

       

                                                                                      

1. The voltage across the 5mF capacitor is?

2. The voltage across the right side circuit (A --> C) is?

3. The equivalent capacitance for the 4 microfarad (mF) & the 2mF parallel capacitors is?

4. The equivalent capacitance for the three capacitors (4mF, 2mF & 3mF) on the right side (Aà C)?

 

5.  What is the difference in electrical potential between points:

            a. A and D

Capacitors in a series all have the same charge. It takes twice the voltage to keep its charge in the 3mF as the two capacitors in parallel (6mF as compared to the 3mF capacitor). Total voltage from A to C is 100V. So,  

1/3 of 100V = 33.3V  (from A to D)

   

b. D and C

 

6. The equivalent capacitance for this network is most nearly?

 

7.  A) The charge stored in the 5‑microfarad capacitor?

Q = CV

      B) The charge in the 4mF capacitor?

      C) The charge in the 2mF capacitor?

       D) The charge in the 3mF capacitor?

 

8. What is the PE of the 5‑microfarad capacitor? 

1/2(CV2)

 

9. A parallel-plate capacitor has a capacitance C1.  A second parallel-plate capacitor (same dielectric constant Eo) has plates with 10 times the area and 5 times the separation.  The capacitance of the second capacitor (C2) is most nearly

C1 = EoA1/d1

C2 = Eo10A1/5d1

 

10. Ten Joules of work is needed to move 2 coulombs of charge from one point to another (1m) with no change in velocity. 

a.  What is the change in electrical potential?

 W = D1/2qV = 1/2 CV2 

    b. What is the change in potential energy?   

      W = DPE = q DV 

c.  What is the electric field strength?

E = volts per meter (V/m)

 

 

 

 

 

 

 

 

11. Two large parallel conducting plates -Q and Q are connected to a battery of emf  

       12V, as shown above.

a. Does the force at various distances between the plates?

b. Does the potential at various distances between the plates?  If so, where is it the highest & lowest potential?

 

 

12.  

  

 

 

 

 

 

 

 

 

12. Three identical resistors (10W), and a capacitor of 1.0 x 10-9 F are connected to

a 30 V battery, as shown in the circuit diagram above. Switches S1 and S2 are initially closed, and switch S3 is initially open. A voltmeter is connected as shown. Find:

a. Current through the battery

b. Determine the reading on the voltmeter.

c. Switches S1 and S2 are now opened, and then switch S3 is closed. Determine the charge Q on the capacitor after S3 has been closed for a long time.