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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

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
Coulomb’s Law – The 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
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
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).


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
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

properties of the electric potential

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

capacitors or condensers

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

  • 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- (-)



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?

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.



Farad (F)

determination of capacitance

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
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 11
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

example 24 3

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
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


  • 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 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
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

capacitors in series

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!!!