Although related, voltage ( V ) and power ( P ) are fundamentally different entities. The voltage within an electrical system is a measure of how many joules ( J ) of energy each coulomb ( C ) of charge ( q ) carries with it. Power is a measure of the rate at which energy is produced or consumed within such systems. It is important, however, not to confuse the terms “ electron volt ( eV ) “ and “ kilowatt hour ( kWh ) “ with voltage and power; these terms are used to describe standalone quantities of energy. Fortunately, mathematical derivations of the eV and kWh are relatively straightforward. First, an electron volt is the amount of kinetic energy ( KE ) gained by an electron that is accelerated through a potential difference of 1 volt within a vacuum:
1V = 1J / C
q = ( C )
1 eV = ( 1.602 x 10-19 C )( 1J / C ) = 1.602 x 10-19 J
When the energy potential traversed by an electron is established by the parallel plates of a capacitor, the electron in question will travel from the negatively charged surface towards the positively charged one via an electric field ( E ):
An electric field is the force per unit charge experienced by charges within the field. E-fields emanate from positive charges and terminate on negative charges; thus, an electric field will repulse a positive charge and attract a negative charge towards it. The distance ( d ) separating the plates of a capacitor provide a means of determining the value of the potential difference across them:
E = F /q
U ( Work ) = Fd
*** The usage of “ U “ here to represent energy is done to avoid confusion. ***
Ed = ( Fd / q ) = ( U / q ) = Vc
Thus, the energy quantum contained by a charge that accelerates across the E-field is as follows:
U = qVc = qEd
An immense and wide-ranging array of electrical imaging devices rely upon the ability to accelerate charges across an electric field en-route to a collision with a pixelated screen:
Q: How many electron volts per second of energy are supplied by a 100 W lightbulb?
A: 100 W = 100 J / s
( 100 J / s )( 1 eV / 1.602 x 10-19 J ) = 6.24 x 1020 eV / s
Q: An electron is released from rest within a uniform electric field that has a magnitude of 2.00 x 104 N/C. At what rate will the electron accelerate during its time-of-travel?
A: F = ma
a = ( F / m )
F = qeE
F = ( 1.602 x 10-19 C )( 2.00 x 104 N/C ) = 3.20 x 10-15 N
a = ( 3.20 x 10-15 N / 9.11 x 10-31 kg ) = 3.52 x 1015 m/s2
RENAISSANCE PHYSICS 