Capacitors (0430)

Charge on Capacitors

We set up a closed circuit to charge a capacitor. We then used the charged capacitor to light a bulb without the batteries in the circuit. We measured how long it took for the bulb to go completely unlit and graphed our results.

Capacitor discharge.

Voltage cross fully charge the capacitor.

In this photo, we draw a sketch of the brightness of the bulb and the time when it is placed across a charged capacitor without the battery present. 

It is a non-line graph. at first, the time is 0, the bulb is most brightness, then as time increases, the bulb gets dimer and dimer.

Here, we tried to obtain the 5uF capacitor by combining five 1uF capacitor in parallel.

 

The red curve represents the charging capacitor, voltage goes up over the course of time. While the blue curve represents the discharging the capacitor, the voltage drops over the course of time. The equation we used for the curve fit is y=A^(1-exp(-Ct))+B for both curves. Interestingly, the C are the same for both graphs, which implying that C is a constant, we found that C is actually 1/Tau=1/RC (R, C are constants, therefore Tau has to be constant)

Here we were trying to find the unit of Tau, to prove that Tau is actually R*C.

After we found the unit of Tau, we set up the equation Q/C= I*R. Since I=dq/dt. We can integrate both side dq/q=dt/Tau. The result is the last equation we wrote on the board.

We said that brightness can be understood as power. Since I=dq/dt=d(CV)/dt)= C*dV/dt. We can see that I graph would be the derivative of voltage graph

Finding Q max here. since Q=V*C= 4.5*e^(-t*C). Q will be max  when t=0, which then equals to 0.45 C

Summary:

     In today’s class, we analyze the idea of capacitors and the charge buildup and discharging of capacitors in a circuit. We did many experiments on capacitors to understand how they work, and how they affect a circuit. We know that what happens when connected in series or parallel, and we did experiments at the different types of RC circuits about developing a relationship between their charging and discharging.  At the end of the class we did some excises to calculate capacitance, time constants, and maximum charge of a circuit 

Capacitors and Capacitance (0428)

Capacitors and Capacitance

We did a quiz about the Kirchhoff's law.

We use Kirchhoff's Current Law and Kirchhoff Voltage Laws to find 3 equations to find the voltage, current and power of each line, and find the total power of them.

We begin to learn Capacitance. it can store charges, the unit is F, the symbol is C.In this photo, Q is charges and V is voltage, and we know the equation Q=VC, and we can find the work done by capacitance is W= 1/2*V^2*C

We plotted our data point using excel. The graph turned out to be 1/x curve.

K here is the dielectric constant. The formula Q=K*epsilon_not *E*A derived from E=Q/epsilon_not*A (Gaussian's surface chapter). The K constant represents different materials will behave differently in magnitude.

Here is some assigned problem. We were asked to find the side of the rectangular plate. Information needed was given.

We were given two capacitors. We measured capacitance of each capacitors individually. Then we obtained the results of capacitors in series and parallel circuit. Interestingly, to find the equivalent capacitors, we would have to do the opposite process to find the equivalent resistance

Our work to find equivalent capacitor. First,we noticed that top two capacitors were in series, so we combined them. Then we combined the two bottom ones. Now, we had a total of 3 parallel capacitors. we just needed to add them up.

The 3 circuits are parallel to each other. We first find the combined the two capacitors of the top ones, which were in series. Then we combined that capacitor to the bottom one, which were parallel. We redrew our circuit which just a single battery and a single equivalent capacitor. We then find the Q1= C1*V1. Q1=Q2. Therefore, C1V1= C2*V2. We now have V2 and V1. V3= V1+V2.=Vb

Electric Circuit, Kirchhoff's Law (0421)

Electric Circuit, Kirchhoff's Law

The image show above is electrical circuit. Also porf.Mason asked the brightness of the light bulbs when the switch was closed. The light bulbs theoretically had the same brightness. However, since the lower light bulb was defective, the brightness did not come out as we expect

That is our prediction. Because the voltage difference between two side of bulb is zero, so the bulb doesn't light.

Theoretically, the two bulb have same brightness, because the voltage difference cross the bulb is same.

That's our answer.

That's our set up.

We were filling the table on the lab manual. Our jobs were to find the potential difference for each set up

We using the color band code compare measurement, the measurement almost in the range.

Here we were trying to simplify the combination of resistors by making them as like one equivalent resistor. 

For parallel: R_eq= (1/R1+1/R2+...1/Rn)^-1

For series: R_eq= R1+R2+..+Rn

Kirchhoff's Law applying.

Today we went back to study electric circuit. This time we emphasized on theoretical side rather than just knowing how to set the wire up as we did in the last lab. We learned about the Kirchhoff's rule in circuit along with the process of how to equivalent resistors (we believe that we can apply the same process to find the intensity also).

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.