Electrostatics - Potential: The Definition Of Voltage

Discussion in 'Physics' started by Nirvana, Aug 21, 2006.

  1. Nirvana

    Thread Starter Well-Known Member

    Jan 18, 2005
    In the previous discussions of Electrostatics Coulomb's Law, Electric Field Strength and Electric Flux Density were all defined.
    Here with the knowledge of the above, the definition of voltage will be defined in terms of potential and energy.
    Potential is the quantity of energy which exists between a place of higher energy to a place of lower energy. Using this definition positive and negative work shall be discussed. Positive work is work which is done against the system (increasing the potential). Conversely negative work is the work done by the system (decreasing the potential).
    In terms of Electrostatics then, if we take two charges of the same polarity - say positive, and we fix one charge in position and allow the other charge free to move. Let Q1 be the fixed charge and Q2 be the moveable charge, then as we bring Q2 closer and closer to Q1 we are going against the system as we are going against the force of repulsion due to the like charges. As we get closer to the fixed charge Q1 we are increasing the potential (positive work). If we then release charge Q2 it would move away from Q1 therefore causing a decrease in potential (negative work).
    In terms of force (F), the force will be equal to the energy taken to move this charge Q2 per distance away from Q1. Expressing this mathematically this becomes:
    δW/dl = F. So to calculate the amount of energy we multiply both sides of the formula by dl to get; δW = F.dl, now recalling the definition of force from Coulomb's Law F = QE where E is the electric field strength. In order to calculate the total energy (W) we need to integrate QE where all the steps are shown;
    W = ∫-QE (Q is a constant remember i.e. its value doesn't change, we have chosen it, so we take Q out as a constant). Therefore we have: W =- Q2 ∫ E . dl remembering the formula for E we get:W = -Q2 ∫ Q1/4πεol2 . dl, NOTE: we have - Q as when we release the charge we experience negative work/potential energy. Taking out constants again, we get. W = -Q2 * Q1/4πεo ∫1/l2 . dl, carrying on the integration will give the overall energy (W), a more useful term would be the energy spent per unit charge or rather energy per Q2 - so divide throught by Q2 to get. W/Q2 = - Q1/4πεo ∫1/l2 . dl which gives: - Q1/4πεo ∫l-2
    W/Q2 = - Q1/4πεo [ l-1/-l] = - Q1/4πεo [- 1/l], now in a system say between two points a and b we use the limits of a and b to get; W/Q2 = - Q1/4πεo [(-1/b) - (-1/a)]
    W/Q2 = - Q1/4πεo [(-1/b) + (1/a)] = Q1/4πεob -Q1/4πεoa
    Now the definition of the voltage becomes energy (J) per Coulomb (Q) which we know to be the volt V, hence W/Q2 = Q1/4πεob -Q1/4πεoa = Vba = Vb - Va.
    There we have the definition of Potential and Voltage (V).

    (Please note that squared values can't be shown in the above description due to the writing package. Please use common sense to determine where certain values arise. For example the number 1 looks very much like the letter l in the above descriptions, please do not get confused. Just read it carefully)