Friday July 11. 8:30 AM 9:0 AM Catch up Lecture 3 Slide 5 Electron projected in electric field problem Chapter 23 Problem 29 Cylindrical shell problem surrounding wire Show Faraday Ice Pail no chrage inside shell Lecture 4 Slide 34 Equipotential surfaces, Sharp points
Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=106 m/s.How much is the electron deflected after traveling 1 cm.
The figure belows shows a thin plastic rod of length L and uniform positive charge Q lying on an x axis. Find the electric field at point P1 on the axis, at distance d from one end of the rod. Find the x and y components.
Capacitance is a characteristic of a single isolated conducting object
or a pair of conducting objects or even three objects. To keep it simple, suppose I have sphere and I put some charge on it say an amount
q. Then it will have some voltage V. If now I double the q,
the voltage will double. The point is that the ratio q/V is constant
and it is called the capacitance. C= q/V. This will be true for any combination of objects. Even if there is no charge on the object it still has a capacitance as you will see.
To help you understand capacitance, consider the isolated sphere again in more detail. If I put a chare q on the isolated sphere, the potential at any point r where r>R, the radius of the sphere, is given by the potential for a point charge.
Isolated conducting sphere with total charge uniformly distributed on its surface. Recall we said that the potential for r > R is given by a point charge q at the center
aHow does the capacitance of the charged conducting sphere change when another neutral sphere is nearby.
Notice the redistribution of charge on both spheres. In effect
the potential V of the original sphere is lowered for the same
amount of charge. This means its effective capacitance has increased in the presence of the neutral sphere. Try to demonstrate it.
aWhat is the capacitance of two charged conducting spheres?One of charge +q and one of charge - q.
C = q/V = 40a (1/(1-m))
The effective capacitance increases
for the paired conductor case and that
is why we make capacitors the way
d =20 cm
a =10 cm
m = a/d = 0.5
C= 0.2 x 10-10 F
C= 0.02 nF =20 pF
If d gets very large, then C= 10 pF
for the isolated sphere.
+ + + + + + + + + + +
Area of plate =A
- - - - - - - - - - -
Integrate from - charge to + charge so that
Coulomb/Volt = Farad
Circular parallel plate capacitor
r = 10 cm = 0.1m
A = r2 = (.1m)2
A = .03 m 2
d = 1 mm = .001 m
p = pico = 10-12
Demonstrate that a piece of wire has capacitance by touching electroscope.
Induced charge factored into kappa
Outer metal braid
- to +
Integrate from b to a or - to +
Va is higher than Vb
a = 0.5 mm
b = 2.0 mm
Now if a=0.5mm and b=2.0mm, then
And if k = 2, then
For = 2
0 (for air)
for an isolated sphere
Let b get very large. Then
AlsoElectric Potential Energy of Capacitor
At some point during the charging, we have a charge q on the positive plate.
The potential difference between the plates is
As we transfer an amount dq of positive charge from the negative plate to the positive one, its potential energy increases by an amount dU.
The total potential energy increase is
Find energy density for parallel plate capacitor. When we charge a capacitor we are creating an electric field. We can think of the work done as the energy needed to create that electric field. For the parallel plate capacitor the field is constant throughout, so we can evaluate it in terms of electric field E easily.
volume occupied by E
We are now including dielectric effects:
Electrostatic energy density general result for all geometries.
To get total energy you need to integrate over volume.
R = 6x106 m
The total solar influx is 200 Watts/m2
Only an infinitesimal fraction gets converted to electricity.
Typical electric circuits have several capacitors in them. How do they combine for simple arrangements? Let us consider two in parallel.
We wish to find one equivalent capacitor to replace C1 and C2. Let’s call it C.
The important thing to note is that the voltage across each is the same and equivalent to V. Also note what is the total charge stored by the capacitors? Q.
What is the equivalent capacitor Ceq?
Voltage across each capacitor does not have to be the same.
The charges on each plate have to be equal and opposite in sign by charge conservation.
The total voltage across each pair is:
C1 = 10 F
C2 = 5.0 F
C3 = 4.0 F
a) Find the equivalent capacitance of the three capacitors
C1 and C2 are in series.
C12 and C3 are in parallel.
C1 = 10 F
C2 = 5.0 F
C3 = 4.0 F
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
c) What is the total energy stored in the circuit?
In Fig. 25-28, a potential difference V = 100 V is applied across a
capacitor arrangement with capacitances C1 = 10.0 µF, C2 = 5.00 µF,
and C3 = 15.0 µF. Find the following values.Chapter 25 Problem 28
(b) the potential difference across capacitor 3
(c) the stored energy for capacitor 3
(d)the charge on capacitor 1
(e) the potential difference across capacitor 1
(f) the stored energy for capacitor 1
(g) the charge on capacitor 2
(h) the potential difference across capacitor 2
(i) the stored energy for capacitor 2
An air-filled parallel-plate capacitor has a capacitance of 1.4 pF. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.3 pF. Find the dielectric constant of the wax.
You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in excess of 10,000 V. You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 17 cm tall with an inner radius of 3.1 cm and an outer radius of 3.5 cm. The dielectric strength is 14 kV/mm.
(a) What is the capacitance?
(b) What is the breakdown potential of this capacitor?