lab14--4/16(Potential of continuous charges distribution)

Find electrical potential due to a finite -length line charge

P(10cm,15cm)

length=16cm

density=2.7mc/cm

cut into 16 segments

distance r=sqrt(x^2+y^2)

E=kq/(r)

calculate each segment in spreadsheet and calculate the total

We increment x by 0.01, starting point is the middle of the last segment of the line.

x=(0.1-0.09)/2+0.09=0.095

x=(0.09=0.08)/2+0.08=0.085

x=(0.08-0.07)/2+0.07=0.075

......

Then we plug in the equation,E=k*dQ/sqrt(x^2+y^2) by increment x

total E=2.46E6

We can also use integral to evaluate the finite length charged E field:

Activity: Calculation of the Potential from a Charged Ring

Then, we derive the electrical field at point p from the charged ring E=kQx/(x^2+a^2)^1.5

Activity:  Sketches of Electric Field Lines and Equipotentials

1.Suppose that you are a test charge and you start moving at some distance from the charge below (such as 4 cm).What path could you move along without doing any work—that is,Eds• is always zero?  What is the shape of the equipotential surface?  Remember that in general you can move in three dimensions.(left top corner on white board）

2.Find some equipotential surfaces for the charge configuration shown below, which consists of two charged metal plates placed parallel to each other.  What is the shape of the equipotential surfaces?（right top corner)

3. Find some equipotential surfaces for the electric dipole charge configuration shown below.(bottom left)

Electric Potential Lab/Activity

Questions:

How much work would you have to do to move 1 coulomb of positive charge from

a) x = 0 cm to x =3 cm? _____4.2E-19____

b) x = 4 cm to x =6 cm? ______0.44E-20___

c) x = 5 cm to x =2 cm?? ______2.7E-19___

How much work would you have to do to move 32 mC of charge from

a) x = 0 cm to x =3 cm?? _________

W=q(V2-V1)=32*10E-6*(0.23+1.55)=5.60E-5 J

b) x = 4 cm to x =6 cm? _________

W=q(V2-V1)=32*10E-6*(0.37+0.22)=1.88E-5 J

c) x = 5 cm to x =2 cm? _________

W=q(V2-V1)=32*10E-6*(0.37+0.44+0.9)=1.44E-3 J

How much work would you have to do to move –25 millicoulombs of charge from

a) x = 0 cm to x =3 cm?? _________

W=q(V2-V1)=-25*10E-3*(0.23+1.55)=-0.0445 J

b) x = 4 cm to x =6 cm? _________

W=q(V2-V1)=-25*10E-3*(0.37+0.22)=-0.01475 J

c) x = 5 cm to x = 2 cm? _________

W=q(V2-V1)=-25*10E-3*(0.23+1.55)=-0.04275 J

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Take your Position and Potential columns copy them to LoggerPro.

Make a graph of Potential (y-axis) vs. Position (x-axis, in meters).