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PHYS 1444 Lecture #5. Tuesday June 19, 2012 Dr. Andrew Brandt. Short review Chapter 24 Capacitors and Capacitance. Coulomb’s Law – The Formula. Formula. A vector quantity. Newtons Direction of electric (Coulomb) force (Newtons) is always along the line joining the two objects.

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### PHYS 1444 Lecture #5

Tuesday June 19, 2012

Dr. Andrew Brandt

• Short review
• Chapter 24
• Capacitors and Capacitance

PHYS 1444 Dr. Andrew Brandt

Coulomb’s Law – The Formula

Formula

A vector quantity. Newtons

• Direction of electric (Coulomb) force (Newtons) is always along the line joining the two objects.
• Unit of charge is called Coulomb, C, in SI.
• Elementary charge, the smallest charge, is that of an electron: -e where

PHYS 1444 Dr. Andrew Brandt

Vector Problems
• Calculate magnitude of vectors (Ex. force using Coulomb’s Law)
• Split vectors into x and y components and add these separately, using diagram to help determine sign
• Calculate magnitude of resultant |F|=(Fx2+Fy2)
• Use = tan-1(Fy/Fx) to get angle

PHYS 1444 Dr. Andrew Brandt

Gauss’ Law
• The precise relation between flux and the enclosed charge is given by Gauss’ Law
• e0 is the permittivity of free space in the Coulomb’s law
• A few important points on Gauss’ Law
• Freedom to choose surface
• Distribution of charges inside surface does not matter only total charge
• Charges outside the surface do not contribute to Qencl.

PHYS 1444 Dr. Andrew Brandt

Example 22-3: Spherical conductor.

A thin spherical shell of radius r0 possesses a total net charge Q that is uniformly distributed on it. Determine the electric field at points (a) outside the shell, and (b) within the shell. (c) What if the conductor were a solid sphere? *q3

Figure 22-11. Cross-sectional drawing of a thin spherical shell of radius r0 carrying a net charge Q uniformly distributed. A1 and A2 represent two gaussian surfaces we use to determine Example 22–3.

Solution: a. The gaussian surface A1, outside the shell, encloses the charge Q. We know the field must be radial, so E = Q/(4πε0r2).

b. The gaussian surface A2, inside the shell, encloses no charge; therefore the field must be zero.

c. All the excess charge on a conductor resides on its surface, so these answers hold for a solid sphere as well.

Key to these questions is how much charge is enclosed

PHYS 1444 Dr. Andrew Brandt

Example 22-4: Solid sphere of charge.

An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0. Determine the electric field (a) outside the sphere (r > r0) and (b) inside the sphere (r < r0).

*q4

Solution: a. Outside the sphere, a gaussian surface encloses the total charge Q. Therefore, E = Q/(4πε0r2).

b. Within the sphere, a spherical gaussian surface encloses a fraction of the charge Qr3/r03 (the ratio of the volumes, as the charge density is constant). Integrating and solving for the field gives E = Qr/(4πε0r03).

PHYS 1444 Dr. Andrew Brandt

Example 22-5: Nonuniformly charged solid sphere.

Suppose the charge density of a solid sphere is given by ρE = αr2, where α is a constant. (a) Find α in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.

Solution: a. Consider the sphere to be made of a series of spherical shells, each of radius r and thickness dr. The volume of each is dV = 4πr2dr. To find the total charge: Q = ∫ρE dV = 4παr05/5, giving α = 5Q/4πr05.

b. The charge enclosed in a sphere of radius r will be Qr5/r05. Gauss’s law then gives E = Qr3/4πε0r05.

PHYS 1444 Dr. Andrew Brandt

Electric Potential Energy
• Concept of energy is very useful solving mechanical problems
• Conservation of energy makes solving complex problems easier.
• Defined for conservative forces (independent of path)

PHYS 1444 Dr. Andrew Brandt

What are the differences between the electric potential and the electric field?

• Electric potential (U/q)
• Simply add the potential from each of the charges to obtain the total potential from multiple charges, since potential is a scalar quantity
• Electric field (F/q)
• Need vector sums to obtain the total field from multiple charges
• Potential for a positive charge is large near a positive charge and decreases to 0 at large distances.
• Potential for the negative charge is small (large magnitude but negative) near the charge and increases with distance to 0
Properties of the Electric Potential

PHYS 1444 Dr. Andrew Brandt

What is a capacitor?

• A device that can store electric charge without letting the charge flow
• What does it consist of?
• Usually consists of two oppositely charged conducting objects (plates or sheets) placed near each other without touching
• Why can’t they touch each other?
• The charges will neutralize each other
• Can you give some examples?
• Camera flash, surge protectors, computer keyboard, binary circuits…
• How is a capacitor different than a battery?
• Battery provides potential difference by storing energy (usually chemical energy) while the capacitor stores charge but very little energy.
Capacitors (or Condensers)

PHYS 1444 Dr. Andrew Brandt

Capacitors
• A simple capacitor consists of a pair of parallel plates of area A separated by a distance d.
• A cylindrical capacitors are essentially parallel plates wrapped around as a cylinder.
• Symbols for a capacitor and a battery:
• Capacitor -||-
• Battery (+) -|i- (-)

Circuit

Diagram

PHYS 1444 Dr. Andrew Brandt

What do you think will happen if a battery is connected (voltage is applied) to a capacitor?

• The capacitor gets charged quickly, one plate positive and the other negative with an equal amount. of charge
• Each battery terminal, the wires and the plates are conductors. What does this mean?
• All conductors are at the same potential.
• the full battery voltage is applied across the capacitor plates.
• So for a given capacitor, the amount of charge stored in the capacitor is proportional to the potential difference Vba between the plates. How would you write this formula?
• C is a proportionality constant, called capacitance of the device.
• What is the unit?
Capacitors

C is a property of a capacitor so does not depend on Q or V.

PHYS 1444 Dr. Andrew Brandt

Normally use mF or pF.

C/V

or

Farad (F)

C can be determined analytically for capacitors w/ simple geometry and air in between.

• Let’s consider a parallel plate capacitor.
• Plates have area A each and separated by d.
• d is smaller than the length, so E is uniform.
• For parallel plates E=s/e0, where s is the surface charge density.
• E and V are related
• Since we take the integral from the lower potential point a to the higher potential point b along the field line, we obtain
• So from the formula:
• What do you notice?
Determination of Capacitance

C only depends on the area (A) and the separation (d) of the plates and the permittivity of the medium between them.

PHYS 1444 Dr. Andrew Brandt

Example 24 – 1

Capacitor calculations: (a) Calculate the capacitance of a capacitor whose plates are 20 cm x 3.0 cm and are separated by a 1.0 mm air gap. (b) What is the charge on each plate if the capacitor is connected to a 12 V battery? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 1F, given the same air gap.

(a) Using the formula for a parallel plate capacitor, we obtain

(b) From Q=CV, the charge on each plate is

PHYS 1444 Dr. Andrew Brandt

Example 24 – 1

(c) Using the formula for the electric field in two parallel plates

Or, since

we can obtain

(d) Solving the capacitance formula for A, we obtain

Solve for A

About 40% the area of Arlington (256km2).

PHYS 1444 Dr. Andrew Brandt

Spherical capacitor: A spherical capacitor consists of two thin concentric spherical conducting shells, of radius ra and rb, as in the figure. The inner shell carries a uniformly distributed charge Q on its surface and the outer shell an equal but opposite charge –Q. Determine the capacitance of this configuration.

Example 24 – 3

Using Gauss’ law, the electric field outside a uniformly charged conducting sphere is

So the potential difference between a and b is

Thus capacitance is

PHYS 1444 Dr. Andrew Brandt

Capacitor Cont’d
• A single isolated conductor can be said to have a capacitance, C.
• C can still be defined as the ratio of the charge to absolute potential V on the conductor.
• So Q=CV.
• The potential of a single conducting sphere of radius rb can be obtained as

where

• So its capacitance is
• Although it has capacitance, a single conductor is not considered to be a capacitor, as a second nearby charge is required to store charge

PHYS 1444 Dr. Andrew Brandt

Capacitors in Series or Parallel
• Capacitors are used in many electric circuits
• What is an electric circuit?
• A closed path of conductors, usually wires connecting capacitors and other electrical devices, in which
• charges can flow
• there is a voltage source such as a battery
• Capacitors can be connected in various ways.
• In parallel and in Series or in combination

PHYS 1444 Dr. Andrew Brandt

Capacitors in Parallel
• Parallel arrangement provides the same voltage across all the capacitors.
• Left hand plates are at Va and right hand plates are at Vb
• So each capacitor plate acquires charges given by the formula
• Q1=C1V, Q2=C2V, and Q3=C3V
• The total charge Q that must leave battery is then
• Q=Q1+Q2+Q3=V(C1+C2+C3)
• Consider that the three capacitors behave like a single “equivalent” one
• Q=CeqV= V(C1+C2+C3)
• Thus the equivalent capacitance in parallel is

For capacitors in parallel the capacitance is the sum of the individual capacitors

Series arrangement is more “interesting”

• When battery is connected, +Q flows to the left plate of C1 and –Q flows to the right plate of C3
• This induces opposite sign charges on the other plates.
• Since the capacitor in the middle is originally neutral, charges get induced to neutralize the induced charges
Capacitors in Series
• So the charge on each capacitor is the same value, Q. (Same charge)
• Consider that the three capacitors behave like an equivalent one
• Q=CeqV  V=Q/Ceq
• The total voltage V across the three capacitors in series must be equal to the sum of the voltages across each capacitor.
• V=V1+V2+V3=(Q/C1+Q/C2+Q/C3)
• Putting all these together, we obtain:
• V=Q/Ceq=Q(1/C1+1/C2+1/C3)
• Thus the equivalent capacitance is

PHYS 1444 Dr. Andrew Brandt

The total capacitance is smaller than the smallest C!!!