Physics equations/Electrostatics

• Review potential energy and work:

$Work=W=Fd\cos \theta =-\Delta PE$,

where W is work, F is force, d is distance moved, and θ is the angle between the force and the distance moved. PE is the potential energy , which can be used to define electric potential, V:

$PE=qV$,

where q is charge. The units of electric potential is the volt (V). The above equation informs us that 1V=1J/C, where J is the joule and C is the coulomb.

• Reminder: Power = Energy/time...P=dE/dt....1 Watt= 1 Joule/second ... 1 W=1J/s

• The electron volt us a unit of energy equivalent to the potential energy of one electron at one volt of potential energy. Using the charge of the electron, we have,

$1eV=1.602\times 10^{-19}J$

• Homework help

Recall that $KE={\frac {1}{2}}mv^{2}$ is kinetic energy, and that if all forces are associated with potential energy, then energy conservation relates initial to final energy by,

$KE+PE=constant=KE_{i}+PE_{i}=KE_{f}+PE_{f}$

${\frac {1}{2}}mv_{i}^{2}+qV_{i}={\frac {1}{2}}mv_{f}^{2}+qV_{f}$

For example, if a charged particle has zero initial speed, the final speed is given by,

$v_{f}={\sqrt {\frac {2qV_{i}}{m}}}$

This says that if a positively charged particle is at rest at a positive electric potential, then it will be moving at the given speed if it is able to drop down to a region of zero electric potential. Recall that the square root of a negative number does not exist as a real number. If the expression inside the radical (square root sign) is negative, then a particle at rest will not move to the region of zero potential unless an external force acts, just as a golf ball at rest will not roll up a hill.

Potential: We shall find the electric potential by calculating the work to move a charged particle. It is convenient to define the potential at point A relative to the potential at point B using subscripts:

$V_{AB}=V_{A}-V_{B}=-(V_{B}-V_{A})=-\Delta V$.

By the work-energy theorem, potential energy is related to the distance moved and the force associated with the potential field (a concept first introduced using gravitational forces):

$W=-\Delta PE=qV_{AB}$

But we also know that work is related to force and distance:

$W=Fd\cos \theta =qEd\cos \theta$

(since the force of a charged particle in an electric field is F=qE.) If we move the particle parallel to the electric field, θ=0 and cosθ=1. Hence we have for the voltage difference between two points along a uniform electric field:

$V_{AB}=Ed$,

where d is the distance measured parallel to the uniform electric field.The two different equations for electric field (F=qE and V=Ed) inform us that we have two equivalent units for electric field

$1N/C=1V/m$,

where N is Newton, C is Coulomb, V is volt, and m is meter.

Capacitors

Charge separation in a parallel-plate capacitor causes an internal electric field. A dielectric (orange) reduces the field and increases the capacitance.

From https://en.wikipedia.org/w/index.php?title=Capacitor&oldid=598664091

A capacitor consists of two conductors separated by a non-conductive region called the dielectric. Examples of dielectric media are glass, air, paper, and vacuum. In virtually all applications conductors hold equal and opposite charges on their facing surfaces. In SI units, a capacitance of one farad means that one coulomb of charge on each conductor causes a voltage of one volt across the device.

An ideal capacitor is wholly characterized by a constant capacitance C, defined as the ratio of charge ±Q on each conductor to the voltage V between them, so that:

$Q=CV$

The capacitance, C, of a parallel plate capacitor is given by,

$C=\varepsilon _{0}A/d$,

where A is plate area and d is the distance between the plates. In a parallel plate capacitor, the electric field between the plates is uniform, and is given by.

E d = V,

where d is the distance between the plates, E is the electric field, and V is the voltage across the plates. (This is obvious because force times distance is work, and F=qE and qV is the work associated with potential energy.) More complicated formulas for the electric field are available for other types of capacitors, such as the co-axial capacitor. Two ways to calculate the energy stored in a capacitor yield identical results. One method is to charge a capacitor (theoretically) and calculate the power. The other way is to (theoretically) place two plates in contact, and then do work to separate the charged plates, which are attracted to each other. The details of both calculations are beyond the scope of this class. But, and interesting result emerges: The result is consistent with the concept that electric field energy density (Joules per cubic meter) equals:

${\frac {energy}{volume}}={\frac {\varepsilon _{0}}{2}}|{\vec {E}}|^{2}$

where ${\vec {E}}$ is the electric field. Also, the energy of a capacitor is given by,

$U={\frac {1}{2}}QV={\frac {1}{2}}CV^{2}={\frac {1}{2}}{\frac {Q^{2}}{C}}$

Another important type of field energy is magnetic field energy. The fact that these fields possess energy allows light to transport energy.