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Voltage confusion

  1. Sep 9, 2005 #1
    voltage confusion!!!!

    Ok, I am an EE major that is confused as Hell with the concept of voltage, and I need some assistance, since understanding voltage is extremely important to my major. I am going to explain my understanding of many many fundamental concepts not because I don't think you know what they are, but because I feel it is essential to explaining voltage overall, and a misconception of one of the fundamentals could be throwing me off somewhere and causing my confusion.

    In order to truly understand what voltage is physically, one must understand charge, force (specifically, coulomb force), electric field, and energy (specifically, potential energy) and each of these will be explained evne though you are already experts on the subject matter, so bear with me.

    First of all, charge is a fundamental physical quantity that cannot be created nor destroyed and accounts for the electrostatic interaction between objects. Charge comes in two types: positive and negative. Two types of charge are needed to account for the TWO types of interaction: attraction and repulsion. Electrons have negative charge and protons have positive charge, so a transfer of electrons between two objects accounts for charge imbalances between protons and electrons, causing opposite charges. Excess of electrons in an objects means negative charge and excess of protons in an object means positive charge. This is very basic but I'd like to put this into my own words.

    Second, force needs to be understood. force is determined to be mass*acceleration, which means that a NET force is only applied if a mass is being accelerated. If there is a constant velocity or no motion at all, when a force is applied, then there must be another force cancelling it, such as friction, or a wall exerting a force back. Electrostatic force is a mutual force between two charged objects. This force can be attractive or repulsive. Coulomb's law is a mathematical model for electrostatic force, which is F = [k(Q1)(Q2)]/R^2, where Q1 and A2 are the two charges involved and R is the distance between the two charges.

    Electric field is a medium that is used as a model to explain action at a distance in electrostatic interaction. It flows from positive charge to negative charge. Mathematically, electric field is the force a charge could exert on a POSITIVE test charge per unit charge of the test charge (yes, poor wording here). Since the test charge is always positive, the sign of the e-field producing charge determines the direction of the e-field, and thus the direction of the force. Thus, a positive charge produces a positive e-field and a positive force (repulsive) on a positive test charge, and a negative charge produces a negative e-field and a negative force (attractive) on a positive test charge. Basically, repulsion is a positive electrostatic force and attraction is a negative electrostatic force. *whew*

    Energy is defined as the capacity to do work. Work is simply a force being exerted through a distance (word = force*distance). The unit for energy is the joule. There are two types of energy: potential energy and kinetic energy. An object has potential energy if a force CAN be exerted through a distance if the force creating an equilibrium environment is taken away. Kinetic energy is another type of energy, and an object with kinetic energy can also do work, or exert a force through a distance. KE = (0.5)mv^2. A mass moving at a constant velocity (0 force on the object moving) has the capacity to exert a force on another mass through a certain distance. The potential energy lost by an object is equal to the kinetic energy gained by an object.

    Now, voltage can finally be explained. Voltage is basically the work per unit charge that a given charge can do between a certain distance if it is placed in an electric field and has the freedom to move. This is easy to understand. Since electric-field is simply the force per unit charge we sinple need to multiply by the distance. However, since electric-field varies inversely as the square of the distance, we cannot simply multiply, and we must integrate. So, voltage is minus the integral of E-field times dR. OK, simple. Now, say we would like to determine the voltage across the terminals of a battery. Since the terminals are opposite charges, the e-field is going to be negative, or -kQ/R^2. Integrating e-field with respect to R gives (kQ)/R + C. Since we must multiply by a -1 we get (-kQ)/R + C. Integrating between the two opposite charges (between 0 and R, since the distance between the negative charge and itself is 0, in a sense, and the distance between the negative charge and the positive charge is R) we get [(-kQ)/R] - [(-kQ)/0] = [(-kQ)/R] + [(kQ)/0]. Obviously, this is a problem, since we get a result of infinity. What am I not understanding about the voltage of a battery? In summation, what is the voltage between two charges, such as a battery. I am clearly thinking of the entire concept of the measurement of voltage incorrectly since the integral of the e-field vs. distance graph between a distance of 0 and a distance of R is infinity, since e-field asymptotically approaches x = 0 so the integral is infinty. I think the concept of a GROUND (arbitrary reference point assumed to have a voltage of zero, in most cases earth ground is used) has to fit in there somewhere but I don't understand where. HELP ME!!!

    I know this is a mouthful, but I NEED some help. I think I would be ok if I just seeked a superficial understanding of voltage across two charges, but I am not that type of person.

    I think this confusion is causing me to really be confused with circuits in general and it is making me sad.

    THANKS A TON FOR YOUR HELP!!
     
    Last edited: Sep 9, 2005
  2. jcsd
  3. Sep 9, 2005 #2
    Also, can someone explain the meaning of the sign on a voltage? I assume that the sign is negative simply because the e-field is negative when the charge emitting the e-field is negative, which would in turn translate the e-field vs. distance graph about the x-axis, causing a negative voltage, correct? So, negative e-field emitting charge and a positive test charge means a negative voltage, and a positive e-field emitting charge and a positive test charge means a positive voltage. Thus, attractive force means negative voltage and repulsive force means positive voltage, right?
     
    Last edited: Sep 9, 2005
  4. Sep 10, 2005 #3
    Ok, let me restate my problem in a more simple way. I am trying to determine the voltage at all points due to a positive point charge, for instance. so I am going to integrate -integral(E*dR). Integration is needed to find the total work a positive test charge can do per unit charge because e-field varies inversely with the square of R, so I cannot simply multiply E by R. However, e-field varies inversely with the square of distance, so its integral from 0 (distance from reference positive charge) to R (distance to positive test charge) is INFINITY. What am I doing wrong???? HELP ME PLEASE!!

    I've looked for some help on the internet, but every website simply MULTIPLIES e-field by distance to get voltage, but obvious you cannot do that, since e-field varies inversely with the square of the distance and you must integrate.

    Obviously, I must be misunderstanding the meaning of a battery's voltage wrong, because by this logic, the voltage of any battery is infinity.
     
  5. Sep 11, 2005 #4

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    Using your own concepts:
    [tex]E = \frac{K}{R^2}[/tex]
    [tex]\int_{\epsilon}^{r}EdR = \int_{\epsilon}^{r}\frac{K dR}{R^2} = \frac{K}{r} - \frac{K}{\epsilon}[/tex]
    Of course the voltage at R = 0 would be infinity. In any other point of the space it is inversely proportional to the distance to the charge.
     
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