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.