Electric Charge and Electric Field

Review Slide 1 - The Atom

• A model of a neutral Calcium-40 atom

• An isotope will have a different number of Neutrons

• An ion will have a different number of Electrons

• A different element will have a different number of Protons

  • Electrons and Neutrons can very too, but proton number is what is important for identifying the element.

Review Slide 2 - Subatomic Particles

• Relative mass is in AMUs
• Relative charge is in Elementary Charge Units (e)
• All particles of the same type are identical there is no variance in charge or mass.

Review Slide 3 - Proton and Neutron Structure

• Up Quarks (u) have charge of \(+\frac{2}{3}\)e, Down Quarks (d) have charge of \(-\frac{1}{3}\)e.

  • The Proton charge is: (+2/3 + 2/3 - 1/3)e = +1e
  • The Neutron charge is: (+2/3 - 1/3 -1/3)e = 0e
  • Neutrons are electric dipoles and can produce electromagnetic effects.

• The Color Code: Red, Green, Blue, is how we account for the Nuclear Strong Force.

  • Three things need to add to zero so red,green, and blue, when mixed, make white.
  • This is all I’ll say of the Nuclear Strong Force in this class.

Review Slide 4 - Charge Units

• The elementary charge unit (e) is set to equal the magnitude of charge of an electron.

  • The electron has charge -1e
  • The proton has charge +1e
  • This is the smallest charge used in chemistry
  • Use this charge unit for atmomic and molecular systems

• The Coulomb is a unit of charge that is equal to \(6.2\times10^{18}\)e.

  • This is the unit of charge to use in laboratory and other large systems

Direction of Eletrical Force

• positive-positive charges repel
• negative-negative charges repel
• positive-negative charges attract

Review Slide 5 - Coloumb’s Law

• The magnitude of electric force, for point charges ,is determined by Coulomb’s Law:

\[ F_e = \frac{1}{4\pi \epsilon_o} * \frac{|q_1q_2|}{r^2} \]

• \(q_1\) and \(q_2\) are the quantities of electric charge in Coulombs (C)
• r is the distance between charges in meters (m)

• \(\epsilon_o\) is the electric constant in vacuum: \(8.854 \times10^{-12} \frac{C^2}{N m^2}\)

  • This is nature’s exchange rate for Charges and distance to force.

• In High School texts they often let \(k = \frac{1}{4\pi \epsilon_o} = 8.988 \times 10^9 \frac{N m^2}{C^2}\)

• Newton’s Law of Universal Gravitation is algebraically similar to Coulomb’s Law, but \(G = 6.67\times10^{-11} \frac{N m^2}{kg^2}\) is much smaller.

• This indicates that the Electrostatic Force is much stronger than Gravity.

The Electric Field

• According to Coulomb’s Law electrostatic forces act at a distance through a vacuum.

  • Note that the Gravitational Force acts the same way.

• The medium that transmits the electrostatic force is called the Electric Field (\(\vec{E}\)).

  • Note that we also discussed the Gravitational field () last semister.

\[ \vec{E} = \frac{\vec{F_e}}{q} \]

• \(\vec{E}\) is a vector that will point in the same direction as \(\vec{F_e}\).

• Units of \(\vec{E}\) are \(\frac{N}{C}\)

• q is called the test charge. Conventionally is is assumed to be a positive charge.

• What Electric Field tells you is how much electrostatic force to expect at a point in space for an arbitrary positive charge.

• A negative charge in an electric field will yield the same magnitude of force, but opposite direction.

Question 1

1. A charge of +2 C is placed in an electric field at a point where \(\vec{E} = +5 \frac{N}{C} i + -2 \frac{N}{C}j\). What is the electrostatic force the field applies to the charge at that point?

2. What if the charge was in the same field as above, but measured -3 C?

Answers

\[ \vec{F_e} = q\vec{E}\\ \vec{F_e} = +2 C (+5 \frac{N}{C} i + -2 \frac{N}{C}j) = +10 N i + -4 Nj \]

\[ \vec{F_e} = q\vec{E}\\ \vec{F_e} = -3 C (+5 \frac{N}{C} i + -2 \frac{N}{C}j) = -15 N i + +6 Nj \]

Electric Field Lines

• Field lines assume a positive test charge, i.e., the field will give direction of force for a positive charge along the field line.

  • They point away from positive charges
  • They point toward negative charges
• Field lines never intersect.
• The closer the field lines the stronger the field at that point.

• The electric field also permeates the space between the lines.

Field lines are a useful model to get a mental picture of how the electric field behaves.

Coulomb’s Law for an Electric Field

• We can Combine Coulomb’s Law with the definition of electric field to make a formula that will allow you to calculate electric field for a single charge.

\[ \vec{E} = \frac{\vec{F_e}}{q} \\ F_e = \frac{1}{4\pi \epsilon_o} * \frac{|q_1q_2|}{r^2} \\ E = \frac{1}{4\pi \epsilon_o} * \frac{|q_1q_2|}{r^2 q} \\ E = \frac{1}{4\pi \epsilon_o} * \frac{|q_1|}{r^2} \]

• Use the Electric Field Line rules to determine direction to get \(\vec{E}\).

Electric Dipoles

• As we saw from the Phet Simulation, just because a particle has a net charge of zero, does not mean it has zero electrical field.

• If the net charge is due to 2 or 3 charges being being separated, we have an electrical dipole.

• We measure the strength of the electrical dipole by the dipole moment \(\vec{p}\):

\[ \vec{p} = \sum_1^N q_id_i \]

Classical Dipole Moment of a Neutron

Assume that the quarks in a neutron are classical point particles each a distance of \(1\times10^{-15}\) m from the center of the neutron. What would the dipole moment be?

Answer

Using trig, it can be shown that the quarks are assumed to make an equilateral triangle in the neutron with side length equal to: \[ d = 1.73\times 10^{-15} m = 1.73\times 10^{-13} cm \]

to simplify the math, we center the dipole on the up quark such that: \[ |p| = \frac{2}{3}e*0cm + 2*(\frac{1}{3}e* 1.73\times 10^{-13}cm) = 1.15\times10^{-13} e*cm \]

Note:

1. Subatomic particles exist in a state that can be wave-like and be spread out in space. Since this is true, the actual value will be much less than this.

2. Neutrons (as well as protons and electrons) are always spinning, this creates much, much stronger electromagnetic effects. Making this an extremely difficult measurement.

Continued

Our best measurement is:

\[ p_N <= 3.0\times10^{-26} e*cm \]

This is still an active area of research!

Dipoles and E-Fields

• When an electric dipole is placed in an E-field, different electric forces will be applied to the ends of the dipole.

• Neither pole will be set on the center of mass, since the particles carry both the charge and the mass.

• When applying force away from the center of mass, torque may be applied.

• Electric Dipoles will then have a tendency to twist in an electrical field until the torque is zero , i.e., the forces align with the field lines.

\[ \vec{\tau} = \vec{p}\times\vec{E} \\ |\tau| = pEsin(\theta) \]

Example

A dipole of magnitude 1.5 \(\mu C*m\) is in an E-field of 9.0 \(\frac{N}{C}\) at a 60^o angle. What is the magnitude of torque the field applies to the dipole?

Answer

\[ 1.5 \mu C*m = 1.5 \times 10^{-6} C*m \\ |\tau| = pEsin(\theta) \\ |\tau| = 1.5 \times 10^{-6} C*m * 9.0 \frac{N}{C} *sin(60^o) = 1.17\times10^{-5} N*m \]