Potential of continuous distribution(0416)

 Potential of continuous distribution

We were asked to find top and bottom potential of the hollow ring. Find the potential using trig. V=k*delta(q)*cos(theta)/x

Different potential at different location.

Electric field line from a dipole

Using conduct paper to find the electric potential 

Today we learned how to derive a formula using integral to find electric potential of a point by a continuous distribution body. We realized that the negative of the first derivative of electric potential would be electric field ( partial derivative V= - E). We used spreadsheet in many of our derivations. At the end we were given metal plate to find electric potential at different location between two metal collars.

Here is a quick look at the data:

HW:

The graph of potential between the metal collars.

The graph of electric potential on the right side of lower potential collar

Potential increases if we go opposite direction of electric field, and decreases if we go the same direction of electric field

Electric Potential(0414)

Electric Potential

Series set up for dimmest, and parallel set up to give the brightest light

Here is the setup. The data were collected by Logger Pro

measure the changed in temperature vs. time first. Since the temperature was changed, there must be a change in either voltage or potential. The slope is dT/dt. dT=mc*dq and dq=work=VI

Here we say that P=VI=dQ/dt=dT/t*mc. Therefore, double V will have the magnitude of the slope double. This relationship only apply to certain substance

We were given voltage (from the source) and resistance (the slope), we were trying to find the current intensity.

Potential is analogous to Potential energy in 4a physics

Equipotential line for a single charge

Formula for finding potential given distance and charge

Visualization of Vpython code given by Prof. Mason

Our code showing the potential between 3 point charges.

Home Work Code:

from __future__ import division
from visual import*
scene.width =800
scene.heigth=600
##OBSERVATION LOCATIONS:
loc1= sphere(pos=(-.3,.1,0), radius = 0.005,color=color.green)
loc2= sphere(pos=(.2,.1,0), radius = 0.005,color=color.green)
loc3= sphere(pos=(-.2,-.1,0), radius = 0.005,color=color.green)
loc4= sphere(pos=(.1,-.1,0), radius = 0.005,color=color.green)

##CONSTANTS
oofpez =9e9
L=0.8
Q= 5e-9
Nq=400 #number of point charges on the rod
dxq= L/Nq #length/ number of point charge
dQ= Q/Nq
scalefactor =1.0


##display Nq spheres
xq= -L/2 +dxq/2
while xq< L/2:
    sphere (pos =(xq,0.05,0), radius =0.01, color= color.red)
    sphere (pos =(xq,-0.05,0), radius =0.01, color= color.red)
    xq= xq+dxq

##CALCULATIONS
Enet= vector (0,0,0) ## starts out at zero and should build up
xq= -L/2 + dxq/2
while xq<L/2:
## calculate the source  location as a vector:
    sourcelocation= vector (xq,0,0)

##calculate the field dE (a vector) contributed by this point charge source

##add this field contribution dE to Enet:

##move to the next source location:
   
    xq= xq+dxq
print (Enet)
##My calculation
##First point (-.3,.1,0), y1=abs(.1-0.05)=0.05, y2=.1-0.05=.15
y1=0.05
V1=oofpez*Q/L*log (L+sqrt ((L*L+y1*y1))/y1)
y2=.15
V2=oofpez*Q/L*log (L+sqrt ((L**2+y2**2))/y2)
Vnet=V1+V2
L2 = label(pos=loc1.pos,xoffset=20,yoffset=20,text= 'V=%1.2f' % Vnet)

##Second Point (.2,.1,0), y1=abs(.1-0.05)=0.05, y2=.1-0.05=.15
y1=0.05
V1=oofpez*Q/L*log (L+sqrt ((L*L+y1*y1))/y1)
y2=.15
V2=oofpez*Q/L*log (L+sqrt ((L**2+y2**2))/y2)
Vnet=V1+V2
L2 = label(pos=loc2.pos,xoffset=20,yoffset=20,text= 'V=%1.2f' % Vnet)

##Thirdpoint (-.2,-.1,0), y1=abs(-.1-0.05)=.15, y2=abs(-.1+.05)=0.05
y1=0.15
V1=oofpez*Q/L*log (L+sqrt ((L*L+y1*y1))/y1)
y2=.05
V2=oofpez*Q/L*log (L+sqrt ((L**2+y2**2))/y2)
Vnet=V1+V2
L2 = label(pos=loc3.pos,xoffset=20,yoffset=20,text= 'V=%1.2f' % Vnet)

##4th point (.1,-.1,0), y1=abs(-.1-0.05)=.15, y2=abs(-.1+.05)=0.05

y1=0.15
V1=oofpez*Q/L*log (L+sqrt ((L*L+y1*y1))/y1)
y2=.05
V2=oofpez*Q/L*log (L+sqrt ((L**2+y2**2))/y2)
Vnet=V1+V2
L2 = label(pos=loc4.pos,xoffset=20,yoffset=20,text= 'V=%1.2f' % Vnet)

Using the equation V= k*Q/L *ln( (L+sqrt(L^2+y^2))/y) to find the electric potential.

Current, Voltage, Resistance(0409)

Current, Voltage, Resistance

How to light up a bulb?

Draw out two possible arrangements of the set up that would allow the bulb to light up, and also two that does not.

Double batteries 

The electroscope!

Suppose the current would be change depends on where the milliamperes placed.   

Same

Vd =I/pqA

Here were the 2 graphs of potential v. time and current v.time. Notice that both graphs had almost perfectly identical graphs.

We did some practice to find the relationship between the length, radius, and resistance. The resistance is directly proportional to the the length, the longer the wire the higher the resistance. The resistance is inversely proportional to the cross sectional area, the higher the are, the easier for the current to flow, the lower the resistance becomes

Gauss' Law (0331)

Gauss' Law

Charge creator with Faraday's cage. Eventually, we learned that there was no charge within the cage

Finding the unit of K constant between charge and flux

factor relationship between radius and volume (1/3)

Using the relationship between q/Q with v/V to find E

The electrical field of a infinite wire

LOL

Electric Field of Dipole (0326)

Electric Field of Dipole, Torque, Dipole Moment

The work done is the dot product of vector (p) and vector(E)

Electric field lines of two opposite charges

"Electric Field"

Here we have the flux on surface 3 and 4 both equal to zero because they don't have any electric field lines go through.The electric flux of surface 1 is negative because electric field lines enter the surface, electric flux of surface 2 is positive because electric field lines exit the surface. These two flux have the same magnitude with the opposite directions, therefore, the net flux will be zero. Thus, the total flux of the box is zero.

Electric field of two opposite charge