Download presentation

1
**Lecture 8 Capacitance and capacitors**

2
Capacitors Capacitors are devices that store energy in an electric field. Capacitors are used in many every-day applications Heart defibrillators Camera flash units Capacitors are an essential part of electronics. Capacitors can be micro-sized on computer chips or super-sized for high power circuits such as FM radio transmitters.

3
**Definition of Capacitance**

The proportionality constant between the charge q and the electric potential difference V is the capacitance C. The definition of capacitance is The units of capacitance are coulombs per volt. The unit of capacitance has been given the name farad (abbreviated F) named after British physicist Michael Faraday ( ) A farad is a very large capacitance Typically we deal with F (10-6 F), nF (10-9 F), or pF (10-12 F)

4
**The parallel-plate capacitor**

The capacitance of a device depends on the area of the plates and the distance between the plates where A is the area of one of the plates, d is the separation, e0 is a constant (permittivity of free space), e0= 8.85´10-12 C2/N·m2 A +Q d A -Q

5
**the charge on each plate**

Example: A parallel plate capacitor has plates 2.00 m2 in area, separated by a distance of 5.00 mm. A potential difference of 10,000 V is applied across the capacitor. Determine the capacitance the charge on each plate Solution: Since we are dealing with the parallel-plate capacitor, the capacitance can be found as Given: DV=10,000 V A = 2.00 m2 d = 5.00 mm Find: C=? Q=? Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:

6
**Combinations of capacitors**

It is very often that more than one capacitor is used in an electric circuit We would have to learn how to compute the equivalent capacitance of certain combinations of capacitors C2 C5 C1 C2 C3 C4 C1 C3

7
**a. Parallel combination**

Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors, +Q1 -Q1 C1 V=Vab a b +Q2 -Q2 C2 A total charge transferred to the system from the battery is the sum of charges of the two capacitors, By definition, Thus, Ceq would be

8
**Parallel combination: notes**

Analogous formula is true for any number of capacitors, It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors (parallel combination)

9
Example: A 3 mF capacitor and a 6 mF capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q1 -Q1 C1 V=Vab a b +Q2 -Q2 C2 Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

10
Series combination Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors, +Q1 -Q1 C1 +Q2 -Q2 C2 V=Vab a c b A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors, By definition, Thus, Ceq would be

11
**Series combination: notes**

Analogous formula is true for any number of capacitors, It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination (series combination)

12
Example: A 3 mF capacitor and a 6 mF capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited. +Q1 -Q1 C1 +Q2 -Q2 C2 V=Vab a c b Given: V = 18 V C1= 3 mF C2= 6 mF Find: Ceq=? Q=? First determine equivalent capacitance of C1 and C2: Next, determine the charge

13
**Energy stored in a charged capacitor**

Consider a battery connected to a capacitor A battery must do work to move electrons from one plate to the other. The work done to move a small charge q across a voltage V is W = V q. As the charge increases, V increases so the work to bring q increases. Using calculus we find that the energy (U) stored on a capacitor is given by: V V q Q

14
**Example: electric field energy in parallel-plate capacitor**

Find electric field energy density (energy per unit volume) in a parallel-plate capacitor Recall Thus, and so, the energy density is + -

15
**(a) determine the equivalent capacitance of the circuit, **

Example: In the circuit shown V = 48V, C1 = 9mF, C2 = 4mF and C3 = 8mF. (a) determine the equivalent capacitance of the circuit, (b) determine the energy stored in the combination by calculating the energy stored in the equivalent capacitance, C1 C2 V C3 First determine equivalent capacitance of C2 and C3: Given: V = 48 V C1= 9 mF C2= 4 mF C3= 8 mF Find: Ceq=? U=? Next, determine equivalent capacitance of the circuit by noting that C1 and C23 are connected in series The energy stored in the capacitor C123 is then

16
**Capacitors with dielectrics**

A dielectrics is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor Notice that the potential difference decreases (k = V0/V) Since charge stayed the same (Q=Q0) → capacitance increases dielectric constant: k = C/C0 Dielectric constant is a material property +Q -Q Insert a dielectric V0 V

17
**Capacitors with dielectrics - notes**

Capacitance is multiplied by a factor k when the dielectric fills the region between the plates completely E.g., for a parallel-plate capacitor The capacitance is limited from above by the electric discharge that can occur through the dielectric material separating the plates In other words, there exists a maximum of the electric field, sometimes called dielectric strength, that can be produced in the dielectric before it breaks down

18
**Dielectric constants and dielectric strengths of various materials at room temperature**

Dielectric constant, k Dielectric strength (V/m) Vacuum 1.00 -- Air 3 ´106 Water 80 Fused quartz 3.78 9 ´106 For a more complete list, see Table 16.1

19
Example: Take a parallel plate capacitor whose plates have an area of 2 m2 and are separated by a distance of 1cm. The capacitor is charged to an initial voltage of 3 kV and then disconnected from the charging source. An insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to kV. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Given: DV1=3,000 V DV2=1,000 V A = 2.00 m2 d = 0.01 m Find: C=? C0=? Q=? k=? Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as The dielectric constant and the new capacitance are The charge on the capacitor can be found to be

20
**Reorientation of polar molecules**

How does an insulating dielectric material reduce electric fields by producing effective surface charge densities? Reorientation of polar molecules + - + - Induced polarization of non-polar molecules + - Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.

21
**Two capacitors, C1=2mF and C2=16mF, are connected in parallel**

Two capacitors, C1=2mF and C2=16mF, are connected in parallel. What is the value of the equivalent capacitance of the combination? (2) Calculate the equivalent capacitance of the two capacitors in the previous exercise if they are connected in series. (3) A 100pF capacitor is charged to a potential difference of 50V, the charging battery then being disconnected. The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the measured potential difference drops to 35V, what is the capacitance of this second capacitor? (4) A parallel-plate capacitor has circular plates of 8.0cm radius and 1.0mm separation. What charge will appear on the plates if a potential difference of 100V is applied?

22
**(5) In figure 6. 18 the battery supplies 12V**

(5) In figure 6.18 the battery supplies 12V. (a) Find the charge on each capacitor when switch S1 is closed, and (b) when later switch S2 is also closed. Assume C1=1mF, C2=2mF, C3=3mF, and C4=4mF. (14) A 16pF parallel-plate capacitor is charged by a 10V battery. If each plate of the capacitor has an area of 5cm2, what is the energy stored in the capacitor? What is the energy density (energy per unit volume) in the electric field of the capacitor if the plates are separated by air? (15) The energy density in a parallel-plate capacitor is given as 2.1 ´l0-9J/m3. What is the value of the electric field in the region between the plates?

23
**(15) The energy density in a parallel-plate capacitor is given as 2**

(15) The energy density in a parallel-plate capacitor is given as 2.1 ´l0-9J/m3. What is the value of the electric field in the region between the plates? (a) Determine the equivalent capacitance for the capacitors shown in figure 6.20. (b) If they are connected to 12V battery, calculate the potential difference across each capacitor and the charge on each capacitor (7) A 6.0mF capacitor is connected in series with a 4.0mF capacitor and a potential difference of 200 V is applied across the pair. (a) What is the charge on each capacitor? (b) What is the potential difference across each capacitor? (8) Repeat the previous problem for the same two capacitors connected in parallel.

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google