PHYS102 Study Guide

Unit 2: Electrostatics

2a. state Coulomb's law and identify the units of the physical quantities contained in the law

  1. How do two point charges interact with one another? Make a sketch that demonstrates three cases: two positive charges, one positive and one negative charge, and two negative charges. Label all relevant quantities and draw and label force vectors on both charges.
  2. What is the expression for Coulomb's force acting on a charge, in vector form?
  3. How does the sign of the product of two charges determine the direction of the force?
  4. What are the units of the constant  k and the electric permittivity of vacuum  \varepsilon_0 ?

Electric charge is a basic property of the particles that make up matter. There are two types of charge: positive and negative. Electrostatic force, which is the force between charges, is one of the four fundamental forces of nature. Many macroscopic forces that can be easily observed, such as the normal force and friction, are the result of the electric interactions between charged particles on the microscopic level.

The expression for the electrostatic force between two point charges, known as Coulomb's Law, is  \vec{F_{21}}=k\frac{q_1 q_2}{r^2} \widehat{r} . Here,  \vec{F_{21}} is the force on the second charge from the first charge,  k is the electrostatic constant that is equal to  8.99\times 10^9\ \mathrm{N\frac{m^2}{C^2}} ,  r is the distance between the two charges, and  \widehat{r} is the unit vector that points from the first charge toward the second charge. The electrostatic constant can be expressed as  k=\frac{1}{4\pi\varepsilon_0} , where the electric permittivity of vacuum (free space),  \varepsilon_0 , is  8.85\times 10^{-12}\ \mathrm{\frac{C^2}{Nm^2}} . The force on the first charge from the second charge would be the same in magnitude but opposite in direction:  \vec{F_{12}}=-k\frac{q_1 q_2}{r^2}\widehat{r} .

Notice that the sign of the product of the charges determines the direction of the force: when the charges are the same, the product is positive and the force is repulsive; when the charges are different, the product is negative and the force is attractive.

Coulomb's Law is described in Charge and Electric Force (Coulomb's Law).

 

2b. solve problems involving electric forces, electric fields, and electric potentials

  1. What information is necessary to calculate the electric field or potential of a charge distribution? Explain how the vector sum of fields of several charges differs from the scalar sum of potentials of several charges.
  2. What information is necessary to calculate the force on a charge in the electric field or electric potential energy of a charge in the electric field?
  3. What is the relationship between electric field and electric potential?

These are formulae pertaining to problems involving electric forces, fields, and potential:

  • Coulomb's Law:  \vec{F_{21}}=k\frac{q_1 q_2}{r^2}\widehat{r} . This is a force between two point or two spherical charges. In case of the spherical charges,  r is the distance between their centers.
  • The electric field of the point charge, or outside of the uniform spherical charge distribution, is  \vec{E}=k\frac{q}{r^2}\widehat{r} . For a positive charge, it points radially outward; for a negative charge, it points radially inward.
  • The superposition principle states that the electric field of several point charges is a vector sum of the fields of each charge:  \vec{E}=\sum_{i=1}^{n}\vec{E_i} . The force on a point charge placed in this field is  \vec{F}=q\vec{E}=q\sum_{i=1}^{n}\vec{E_i} .
  • Continuous charge distributions are given by charge densities:

    Linear charge distribution, like on a charged string or thin rod, is described by linear charge density,  \lambda . For uniform distributions, the total charge is  Q= \lambda L , where  L is the length of the string or rod. Otherwise, the total charge is the line integral  Q=\int\! \lambda\ \mathrm{d}x .

    Surface charge distribution, like on a charged plane or other surface, is described by surface charge density,  \sigma . For uniform distributions, the total charge is  Q = \sigma A , where  A is the area of the surface. Otherwise, the total charge is the surface integral  Q=\int\! \sigma\ \mathrm{d}A .

    Volume charge distribution, like for a charged solid object, is described by volume charge density,  \rho . For uniform distributions, the total charge is  Q = \rho V , where  V is the volume of the object. Otherwise, the total charge is the volume integral  Q=\int\! \rho\ \mathrm{d}V .
  • The field of a continuous charge distribution at a given point is  \vec{E}=\int \! k \frac{\widehat{r}}{r^2}\ \mathrm{d}q , where  \mathrm{d}q = \lambda\ \mathrm{d}x ,  \mathrm{d}q = \sigma\ \mathrm{d}A , or  \mathrm{d}q = \rho\ \mathrm{d}V for linear, surface, or volume charge distributions, respectively. Here,  r is the distance between each point of the distribution and the given point.
  • Potential at a point in the field of a point charge is  V=k\frac{q}{r} . Potential is a scalar, as opposed to an electric field, which is a vector. Potential at a point in the field of several charges is  V=\sum_{i=1}^{n}k\frac{q_i}{r_i} , and potential at a point in the field of a continuous charge distribution is  V=\int \! k\ \frac{\mathrm{d}q}{r} .
  • Potential energy of one charge in the field of another is  U=k\frac{q_1 q_2}{r} , where  r is the distance between the charges.
  • Electric fields and electric potentials are related as  \vec{E}=-\frac{\mathrm{d}V}{\mathrm{d}\vec{r}} , or  V=\int \! \vec{E}\ \mathrm{d}\vec{r} . For the case of a uniform one-dimensional field, this means  V=-E_x x . The electric field points in the direction of decreasing potential, and electric field lines are perpendicular to the equipotential surfaces at every point.

There are several common types of problems involving electric forces, fields, and potential:

  • Calculating electric field or force on a charge due to several point charges by using superposition. See the examples Electricity.
  • Calculating the electric field of a continuous distribution of charges. Watch Field of Infinite Plane for an example. There is an example of calculating the field of a uniformly charged rod section 22.8 of Light and Matter. This chapter also contains examples of problems involving the relationship between fields and potentials.
  • Calculating electric field using Gauss' Law. See learning outcome 2e below.
  • Calculating electric potential of a distribution of charges, and electric potential energy of a system of charges. See examples 5.3 and 5.4 in Electric Potential.

 

2c. explain Gauss' law in words

  1. How would you illustrate electric field lines, Gaussian surfaces, and electric flux? Draw a few examples, such as field lines of a positive or negative point charge, or field lines of an infinitely large plane.
  2. What is the mathematical expression for electric flux as a surface integral? How is it related to the number of field lines?
  3. How do the number of field lines entering and leaving a Gaussian surface relate to the charge enclosed by the surface?

Gauss' Law demonstrates that a field is inversely proportional to the square of the distance from its source. It applies to electric fields and others, such as gravitational fields. It states that the electric flux, or number of field lines leaving a closed surface, is proportional to the electric charge enclosed by that surface:  \Phi_E=\oint \! \vec{E}\ \mathrm{d}\vec{A}=\frac{Q_{\mathrm{enc}}}{\varepsilon_0} . (Here,  \varepsilon_0=8.85\times 10^{-12}\ \mathrm{\frac{C^2}{Nm^2}} is the electric permittivity of vacuum.) If the electric field is constant on the surface, the flux equals  \Phi_E=EA\cos(\theta) , where  \theta is the angle between the field lines and the surface area vector (the vector perpendicular to the surface with the magnitude equal to its area). This means that if there is no charge enclosed by a surface, the net electric flux through the surface is zero. Since the electric field lines have to begin or end on the charges, if there is no charge enclosed by the surface, no new lines can be created or terminated inside the surface; all lines that enter the surface have to leave it.

Read more about Gauss' Law in section 22.7 of Light and Matter.

 

2d. compare and contrast the electric potential and the electric field

  1. How do the concepts of electric potential energy and electric potential arise from the calculation of work performed by an electric field on a charge placed in the field?
  2. How are field lines and the direction of a field related to the location of its points of equal potential?

The motion of a particle can be described either in terms of the forces acting on it, or in terms of its potential and kinetic energy. In the case of a charged particle placed in an electric field, the force would be an electric force, and its potential energy is the energy associated with the electric field. A force usually described for a charge at one point in time, whereas electric potential and electric fields are more usually indicated when a charge moves from one location to another.

Review the definition of the electric potential by watching Electric Potential Energy, Electric Potential, and Voltage. The relevant relationships and examples are discussed in Electric Potential.

 

2e. solve problems using Gauss' Law

  1. What information must be known about a charge distribution to calculate it using Gauss' Law?
  2. What are the guidelines for drawing a Gaussian surface?
  3. What information is necessary to calculate the electric flux through an enclosed surface?

Gauss' Law can be used to calculate a field in situations where the charge distribution has symmetry, usually spherical or cylindrical. Do not attempt to use derived formulas, such as for the electric field of a sphere, and adjust them to a different problem; these will not work. Always start with Gauss's Law itself, pick an appropriate Gaussian surface, and find the charge enclosed by this surface. When you need to calculate the flux (not field), all you need to know is how much charge is enclosed by the surface in question; the shape of the surface is irrelevant.

Watch Gauss' Law and Application to Conductors and Insulators to review, and see Gauss' Law for some solved examples.

 

2f. solve problems involving the motion of charged particles in an electric field

  1. What is the direction of the force on a charged particle placed in an electric field? How does it depend on the charge of the particle?
  2. Consider a charged particle entering a region with a uniform electric field. What will the particle's acceleration be? Draw the trajectory of a positively charged particle in the field for the cases when the initial velocity is zero, parallel to the field, or perpendicular to the field. What will change if the particle is negatively charged?

The motion of a charged particle in a uniform electric field is similar to the motion of a massive object near the surface of the Earth, which is projectile motion. In both cases, the acceleration is constant. The acceleration of a charged particle in a uniform electric field is  \vec{a}=\frac{q\vec{E}}{m} , where  \vec{E} is the field,  q is the charge, and  m is the mass of the particle. The trajectory is determined by the direction of the initial velocity of the particle. If the initial velocity is zero or parallel to the field, the particle will move in a straight line and be accelerated or decelerated by the electric force. However, if there is an angle between the initial velocity and the field, the trajectory will be parabolic, like the trajectory of a projectile in free fall.

Motion of a Charge in an Electric Field illustrates this effect.

 

2g. define capacitance and describe the factors that determine capacitance

  1. When is a parallel-plate capacitor considered ideal (the electric field is uniform between its plates and zero outside)?
  2. Consider an ideal parallel-plate capacitor with surface charge density  +\sigma on one plate and  -\sigma on another. Assume that the area of the plates and the distance between them is known. What is the electrical potential between the plates? How would you use that to find the capacitance of the parallel-plate capacitor?
  3. If you have two capacitors, how would you connect them to a battery so that they are connected in series? What will the relationship between the charges on each capacitor be in this case? What will the relationship between the voltage on each capacitor and the voltage supplied by the battery be? Use these considerations to determine the equivalent capacitance of two capacitors connected in series.
  4. If you have two capacitors, how would you connect them to the battery so that they are connected in parallel? What will the relationship between the charges on each capacitor be in this case? What will the relationship between the voltage on each capacitor and the voltage supplied by the battery be? Use these considerations to determine the equivalent capacitance of two capacitors connected in parallel.

If a charge is placed on the surface of a conductor or combination of conductors, the resultant electric potential is proportional to that charge:  Q=CV . The proportionality constant  C between the potential and the charge is called capacitance, and it depends only on the geometry of the conductor. For a parallel-plate capacitor,  C=\frac{\varepsilon_0 A}{d} , where  \varepsilon_0 is the electric permittivity of vacuum,  A is the area of the plates, and  d is the distance between the plates. The distance must be much smaller than the size of the plates so that the electric field inside the capacitor is uniform.

Watch Capacitors and Capacitance and Capacitance, and read Charge Storage, Breakdown, and Capacitance to review these concepts.

When two or more capacitors are connected in series:

  • They have the same charge because charge is conserved
  • The sum of voltages on each capacitor equals the voltage supplied by the battery
  • The equivalent capacitance is determined by the formula  \frac{1}{C_\mathrm{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots

When two or more capacitors are connected in parallel:

  • They have the same voltage, which also equals the voltage supplied by the battery
  • The sum of charges on each capacitor equals to total charge, proportional to the voltage supplied by the battery
  • The equivalent capacitance is determined by the formula  C_\mathrm{eq}=C_1+C_2+\cdots

The series and parallel connection of capacitors is discussed in Capacitors in Series and Capacitors in Parallel. Also, see this solved example of determining an equivalent capacitance of a circuit.

 

2h. describe the effect of a dielectric material in a capacitor

  1. What are the differences between conductors and dielectrics? How are charged particles in each material affected when the material is placed in an electric field?
  2. How is the electric field inside a dielectric material related to the external electric field? Considering this, how is the capacitance affected when the dielectric is placed inside a capacitor?

When a dielectric material is placed inside an electric field, the positive and negative charges inside the material experience forces that act in opposite directions. As a result, the molecules inside the dielectric rotate so that the positive charges move toward the direction of the field, while the negative charges move in the direction opposite to the field. The separation of the charges creates the internal field of the dielectric, which is directed opposite to the external field. Therefore, the magnitude of the net field inside the dielectric is smaller than the external field by  K , which is called the dielectric constant of the material. As a result of the decrease in the electric field when a capacitor is filled with a dielectric, its capacitance increases by a factor of  K .

Dielectrics contains the values of dielectric constants for various materials. See the detailed discussion of how capacitors are affected by dielectrics in Dielectrics in Capacitors and this solved example.

 

2i. define electric potential energy and describe how capacitors can be used to store energy

  1. How much work is performed by an external agent to charge a capacitor to a given charge?
  2. What are different ways to express the energy stored by a capacitor using its capacitance, charge, and voltage?

Since work must be performed in order to charge a capacitor, the charged capacitor stores energy equal to that work. The energy stored in a capacitor with the charge  Q is  U=\frac{Q^2}{2C} . Alternatively, since charge is proportional to the voltage as  Q=CV , the stored energy can be expressed as  U=\frac{1}{2}QV=\frac{CV^2}{2} .

Refer to Energy of a Capacitor, Energy Stored by Capacitors, and this solved example.

 

Unit 2 Vocabulary

This vocabulary list includes terms that might help you with the review items above and some terms you should be familiar with to be successful in completing the final exam for the course.

Try to think of the reason why each term is included.

  • Capacitance
  • Capacitor
  • Charge
  • Charge density (linear, surface, and volume)
  • Coulomb's Law
  • Dielectric constant
  • Dielectric material
  • Electrostatic (Coulomb's) force
  • Electric field
  • Electric flux
  • Electric permittivity
  • Electric potential
  • Electric potential energy
  • Field line
  • Gauss' Law
  • Gaussian surface
  • Parallel connection
  • Series connection
  • Superposition principle
  • Voltage