PHY 184

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# PHY 184 - PowerPoint PPT Presentation

PHY 184. Spring 2007 Lecture 13. Title: Capacitors. Notes. Homework Set 3 is done! Homework Set 4 is open and Set 5 opens Thursday morning Midterm 1 will take place in class Thursday, February 8. One 8.5 x 11 inch equation sheet (front and back) is allowed. The exam will cover

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### PHY 184

Spring 2007

Lecture 13

Title: Capacitors

184 Lecture 13

Notes
• Homework Set 3 is done!
• Homework Set 4 is open and Set 5 opens Thursday morning
• Midterm 1 will take place in class Thursday, February 8.
• One 8.5 x 11 inch equation sheet (front and back) is allowed.
• The exam will cover

Chapters 16 - 19

Homework Sets 1 - 4

184 Lecture 13

Review
• The capacitance of a spherical capacitor is
• r1 is the radius of the inner sphere
• r2 is the radius of the outer sphere
• The capacitance of an isolated spherical conductor is
• R is the radius of the sphere

184 Lecture 13

Review (2)
• The equivalent capacitance for n capacitors in parallel is
• The equivalent capacitance for n capacitors in series is

184 Lecture 13

Example - System of Capacitors
• Let’s analyze a system of five capacitors
• If each capacitor has a capacitance of 5 nF, what is the capacitance of this system of capacitors?

184 Lecture 13

System of Capacitors (2)
• We can see that C1 and C2 are in parallel and that C3 is also in parallel with C1 and C2
• We can define C123 = C1 + C2 + C3 = 15 nF
• … and make a new drawing

184 Lecture 13

System of Capacitors (3)

= 3.75 nF

• We can see that C4 and C123 are in series
• We can define
• … and make a new drawing

184 Lecture 13

System of Capacitors (4)

= 8.75 nF

• We can see that C5 and C1234 are in parallel
• We can define
• And make a new drawing

184 Lecture 13

System of Capacitors (5)
• So the equivalent capacitance of our system of capacitors
• More than one half of the total capacitance of this arrangement is provided by C5 alone.
• This result makes it clear that one has to be careful how one arranges capacitors in circuits.

184 Lecture 13

Clicker Question
• Find the equivalent capacitance Ceq

A)

B)

C)

184 Lecture 13

Clicker Question
• Find the equivalent capacitance Ceq

C)

First Step: C1 and C2 are in series

Second Step: C12 and C3 are in parallel

184 Lecture 13

A capacitor stores energy.

Field Theory:

The energy belongs to the electric field.

184 Lecture 13

Energy Stored in Capacitors
• A battery must do work to charge a capacitor.
• We can think of this work as changing the electric potential energy of the capacitor.
• The differential work dW done by a battery with voltage V to put a differential charge dq on a capacitor with capacitance C is
• The total work required to bring the capacitor to its full charge q is
• This work is stored as electric potential energy

184 Lecture 13

Energy Density in Capacitors
• We define the energy density, u, as the electric potential energy per unit volume
• Taking the ideal case of a parallel plate capacitor that has no fringe field, the volume between the plates is the area of each plate times the distance between the plates, Ad
• Inserting our formula for the capacitance of a parallel plate capacitor we find

184 Lecture 13

Energy Density in Capacitors (2)
• Recognizing that V/d is the magnitude of the electric field, E, we obtain an expression for the electric potential energy density for parallel plate capacitor
• This result, which we derived for the parallel plate capacitor, is in fact completely general.
• This equation holds for all electric fields produced in any way
• The formula gives the quantity of electric field energy per unit volume.

184 Lecture 13

Example - isolated conducting sphere
• An isolated conducting sphere whose radius R is 6.85 cm has a charge of q=1.25 nC.

a) How much potential energy is stored in the electric field of the charged conductor?

Key Idea: An isolated sphere has a capacitance of C=4e0R (see previous lecture). The energy U stored in a capacitor depends on the charge and the capacitance according to

… and substituting C=4pe0R gives

184 Lecture 13

Example - isolated conducting sphere

(Why?)

• An isolated conducting sphere whose radius R is 6.85 cm has a charge of q=1.25 nC.

b) What is the field energy density at the surface of the sphere?

Key Idea: The energy density u depends on the magnitude of the electric field E according to

so we must first find the E field at the surface of the sphere. Recall:

184 Lecture 13

Example: Thundercloud
• Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0 km hovers over a flat area, at an altitude of 500 m and carries a charge of 160 C.
• Question 1:
• What is the potential difference betweenthe cloud and the ground?
• Question 2:
• Knowing that lightning strikes requireelectric field strengths of approximately2.5 MV/m, are these conditions sufficientfor a lightning strike?
• Question 3:
• What is the total electrical energy contained in this cloud?

184 Lecture 13

Example: Thundercloud (2)
• Question 1
• We can approximate the cloud-ground system as a parallel plate capacitor whose capacitance is
• The charge carried by the cloud is 160 C, which means that the “plate surface” facing the earth has a charge of 80 C
• 720 million volts

…++++++++++++ …

++++++++++++

184 Lecture 13

Example: Thundercloud (3)
• Question 2
• We know the potential difference between the cloud and ground so we can calculate the electric field
• E is lower than 2.5 MV/m, so no lightning cloud to ground
• May have lightning to radio tower or tree….
• Question 3
• The total energy stored in a parallel place capacitor is
• Enough energy to run a 1500 W hair dryer for more than 5000 hours

184 Lecture 13

Clicker Question
• A 1.0 F capacitor and a 3.0 F capacitor are connected in parallel across a 500 V potential difference V. What is the total energy stored in the capacitors?

A) U=0.5 J

B) U=0.27 J

C) U=1.5 J

D) U=0.02 J

Hint: Use

184 Lecture 13

Clicker Question
• A 1.0 F capacitor and a 3.0 F capacitor are connected in parallel across a 500 V potential difference V. What is the total energy stored in the capacitors?

A) U=0.5 J

U = 0.5 J

184 Lecture 13

Capacitors with Dielectrics
• We have only discussed capacitors with air or vacuum between the plates.
• However, most real-life capacitors have an insulating material, called a dielectric, between the two plates.
• The dielectric serves several purposes:
• Provides a convenient way to maintain mechanical separation between the plates.
• Provides electrical insulation between the plates.
• Allows the capacitor to hold a higher voltage
• Increases the capacitance of the capacitor
• Takes advantage of the molecular structure of the dielectric material

184 Lecture 13

Capacitors with Dielectrics (2)
• Placing a dielectric between the plates of a capacitor increases the capacitance of the capacitor by a numerical factor called the dielectric constant, 
• We can express the capacitance of a capacitor with a dielectric with dielectric constant  between the plates as
• … where Cair is the capacitance of the capacitor without the dielectric
• Placing the dielectric between the plates of the capacitor has the effect of lowering the electric field between the plates and allowing more charge to be stored in the capacitor.

184 Lecture 13

Parallel Plate Capacitor with Dielectric
• Placing a dielectric between the plates of a parallel plate capacitor modifies the electric field as
• The constant 0 is the electric permittivity of free space
• The constant  is the electric permittivity of the dielectric material

184 Lecture 13