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  ### Course: Electrical engineering>Unit 5
  
  Lesson 1: Electric force and electric field
  
  # Electric field
  
  Definition of the electric field. Electric field near a point charge.
  Coulomb's Law describes forces acting at a distance between two charges. We can reformulate the problem by breaking it into two distinct steps, using the concept of an electric field.
  • Think of one charge as producing an electric field everywhere in space.
  • The force on another charge introduced into the electric field of the first, is caused by the electric field at the location of the introduced charge.
  If all charges are static, you get exactly the same answers with electric field as you do using Coulomb's Law. So, is this going to be just an exercise in clever notation? No. The electric field concept comes into its own when charges are allowed to move relative to each other. Experiments show that only by considering the electric field as a property of space that propagates at a finite speed (the speed of light), can we account for the observed forces on charges in relative motion. The electric field concept is also essential for understanding a self-propagating electromagnetic wave such as light. The electric field concept gives us a way to describe how starlight travels through vast distances of empty space to reach our eyes.
  If the idea of force "acting at a distance" in Coulomb's Law seems troublesome, perhaps the idea of force caused by an electric field eases your discomfort somewhat. On the other hand, you might also question if an electric field is any more "real". The reality of an electric field is a topic for philosophers. In any case, real or not, the notion of an electric field turns out to be useful for predicting what happens to charge.
  We introduce electric field initially with static charges to ease into the concept and get practice with the method of analysis.
  
  ## Definition of electric field
  
  The electric field $\stackrel{\to }{E}$ is a vector quantity that exists at every point in space. The electric field at a location indicates the force that would act on a unit positive test charge if placed at that location.
  The electric field is related to the electric force that acts on an arbitrary charge $q$ by,
  $\stackrel{\to }{E}=\frac{\stackrel{\to }{F}}{q}$
  The dimensions of electric field are newtons/coulomb, $\text{N/C}$.
  We can express the electric force in terms of electric field,
  $\stackrel{\to }{F}=q\stackrel{\to }{E}$
  For a positive $q$, the electric field vector points in the same direction as the force vector.
  The equation for electric field is similar to Coulomb's Law. We assign one of the $q$'s in the numerator of Coulomb's Law to play the role of test charge. The other charge(s) in the numerator, ${q}_{i}$, create the electric field we want to study.
  Where $\phantom{\rule{0.167em}{0ex}}\stackrel{^}{{r}_{i}}$ are unit vectors indicating the line between each ${q}_{i}$ and $q$.
  
  ### How to think about electric field
  
  The electric field is normalized electric force. Electric field is the force experienced by a test charge that has a value of $+1$.
  One way to visualize the electric field (this is my mental model): imagined small positive test charge glued to the end of an imaginary stick. (Be sure your imaginary stick doesn't conduct, like wood or plastic). Explore the electric field by holding your test charge in various locations. The test charge will be pushed or pulled by the surrounding charge. The force the test charge experiences (both magnitude and direction), divided by the value of the small test charge, is the electric field vector at that location. Even if you take away the test charge, there is still an electric field at that location.
  
  ## Electric field near an isolated point charge
  
  The electric field around a single isolated point charge, ${q}_{i}$, is given by,
  $\stackrel{\to }{E}=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\frac{{\text{q}}_{i}}{{r}^{2}}\phantom{\rule{0.167em}{0ex}}{\stackrel{^}{r}}_{i}$
  The electric field direction points straight away from a positive point charge, and straight at a negative point charge. The magnitude of the electric field falls off as $1/{r}^{2}$ going away from the point charge.
  
  ## Electric field near multiple point charges
  
  If we have multiple charges scattered about, we express the electric field by summing the fields from each individual ${q}_{i}$,
  $\stackrel{\to }{E}=\frac{1}{4\pi {ϵ}_{0}}\sum _{i}\frac{{\text{q}}_{i}}{{r}^{2}}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{{r}_{i}}$
  The summation is performed as a vector sum.
  
  ## Electric field near distributed charge
  
  If charges are smeared out in a continuous distribution, the summation evolves into an integral.
  $\stackrel{\to }{E}=\frac{1}{4\pi {ϵ}_{0}}\int \frac{\text{d}q}{{r}^{2}}\phantom{\rule{0.167em}{0ex}}\stackrel{^}{r}$
  Where $r$ is the distance between $\text{d}q$ and the location of interest, while $\stackrel{^}{r}$ reminds us the direction of the force is in a direct line between $\text{d}q$ and the location of interest. We will see examples of this integral in two upcoming articles.
  The discussion above defines the electric field. There isn't any new physics, we've just defined some new terms. Now we're ready to move on and use the electric field formulation to analyze two common real-world geometries: the line of charge, and the plane of charge.
  
  ## Want to join the conversation?
  
  • Why electric feild is more stronger then magnetic field and gravitational field ?
  • Hello Rajeshk,
  
  This is an unsatisfying answer but it just is... These are fundamental forces of nature.
  
  As a consolation for my poor answer may I share my favorite video on the topic:
  
  Enjoy,
  
  Aaron
  • what is meant by "If charges are smeared out in a continuous distribution, the summation evolves into an integral."?
  • There are two ways to think about charge. We know that charge is the property of two atomic particles, electrons and protons. This makes it convenient to think about charge as particles, or like a bunch of sand. You can count sand particles (if there are not too many). Coulomb's Law treats charge this way, there's a q1 and a q2.
  
  Another way is to think of charge as a continuous substance, like peanut butter. Peanut butter isn't a collection of particles, it's something different. You charge something by slathering it with peanut butter charge. The charge is uniformly distributed throughout the peanut butter.
  If you see a problem statement like "assume a uniformly charged rod," that's an example of the continuous peanut butter version of charge. Continuous charge will include a density specification like 2 coulombs per meter, or 3 coulombs per cubic inch.
  
  If you are presented with a problem based on peanut butter charge you have to figure how to apply particle-based Coulomb's Law. In this blob of charge we have to somehow identify a charge particle. The trick is to use calculus to focus down on a tiny tiny bit of the charged structure, a bit so small it can be considered a particle.
  
  So in the article you see the equation for the electric field from multiple charges
  
  F = 1/4pieo SUM (q_i/r^2)
  
  In peanut butter charge q_i becomes the differential charge dq, and the SUM turns into (evolves into) an INT (integral).
  
  F = 1/4pieo INT (dq/r^2)
  
  These two equations mean the same thing. In the second we rely on calculus notation to do the bookkeeping for adding up all those infinitesimal dq's.
  • What is an equatorial line?
  • That's a line around the equator of an object. You are probably looking at a problem that includes a sphere, and the sphere has an obvious "up" and "down" axis. The equator is the line half way between the north and south poles of the sphere, just like the Earth.
  • Sir i'm not saying about electric field strength or intensity.I am asking that can we measure how much area an electric field surrounds.?or it is spread to infinity?
  • The strength of the electric field weakens with the square of the distance. So if you double the distance it's only 1/4th of the strength, if you increase the distance ten-fold the strength becomes 1/100th, and so on. Much the same as gravity. So, while it might not be "cut off" at a certain point it obviously diminishes quite fast into trivial strengths. Perhaps we could view it as being "cut-off" when the strength goes below Planck sizes.
  • In the denominator of the equation given in "Electric field near multiple point charges", what is r? Considering we sum over multiple point charges, and our point might not be equidistant from all of them, shouldn't it be r_i?
  • You are correct. The r in the denominator should probably have a subscript of i.
  • electric field is also written in volts/metre. how is it similar to newtons/coulomb and why is volts/metre appropriate.
  • The two representations for electric field are "equivalent", meaning you can derive one from the other.
  
  The volt is the unit of electric potential, which is energy (joules) per unit charge (coulombs). That means 1 volt is a nickname for a potential of 1 joule/coulomb.
  
  One joule of energy is the energy required to accelerate a mass of one kilogram using one newton of force over a distance of one meter. The newton is the unit of force and has units of F = ma or kg*m/sec^2.
  
  So, 1 newton of force is 1 joule of energy applied over 1 meter, or 1 joule/meter.
  
  Imagine you are measuring electric field in the air gap between two capacitor plates. You know the voltage of each plate. V/m would be an appropriate unit to pick. Express volts as joules/coulomb and you get an electric field measurement in joules/(coulomb*meter).
  
  Another way to measure electric field is in terms of newtons per coulomb. You might pick this set of units because you are measuring an electric field by inserting a test charge into the field and measuring the force on the test charge. When you divide newtons expressed as (joules/meter) by coulombs, you get an electric field expressed in joules/(coulomb*meter). That's the same units as volts/meter!
  
  [To create this answer I used Khan Academy's experimental AI tutor, "Khanmigo" to get a starting point. Then I revised the response to (hopefully) make it more clear. This is my first experience using an AI.]
  • how is electrostatics related to transformers?
  (1 vote)
  • Hello Cassiecsy,
  
  Nearly every electronics and physics textbook begins with a discussion of electrostatics. There is talk about charge, charge carriers, and the transfer of charge. These concepts lead an understanding of voltage and current.
  
  Electrostatics helps us understand how we move these charge carriers (electrons in a wire). How to describe the concepts and how to perform measurements.
  
  Since in all cases we are talking about the movement of charges electrostatics has everything to do with transformers.
  
  On the other hand, transformers are electromagnetic devices. Only the vocabulary of electrostatics is used. Unlike capacitors which store energy in electrostatic fields...
  
  Regards,
  
  APD
  • 1. What is 1/4πϵ0? Is it the same as k?
  2. Is Coulomb's Law the same as the electric force equation which is Fe=k(q1q2/r^2)?
  3. Could anyone explain where [ F = (1/4πϵ0)(Q times Qi/r^2)(ri) ]?
  (1 vote)
  • 1. Yes. 1/4pie_o = k. It is just two forms of an arbitrary constant. The 1/4pi comes from a theory you study later in electromagnetics, Gauss's Theory, which deals with the surface area of a sphere (that's where 1/4pi comes from). Since Gauss's Theory is so important and is naturally written with the 1/4pi notation, some teachers bring that form all the way back to Coulomb's Law.
  
  2. Yes. Coulomb's Law IS the electric force equation.
  
  3. Coulomb's Law arises from a real-world experiment. It is not something you can derive from first principles. Coulomb himself designed and performed the experiment to measure electric force, and used the data he recorded to create the Law named for him. The little ri at the end is a vector notation to indicate the force lies along the line between the two charges.
  • how fast is the speed of light?
  (1 vote)
  • Simple answer, it is 3 x 10^8 m/s or 1.07 billion km/h.
  
  To give you a picture, light takes 8 minutes to travel from sun to earth. On the other hand, the fastest bullet train on earth (maximum speed being 460 km/h) takes about 7.7 million years!
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