ben gurion university of the negev lecturer: daniel rohrlich teaching assistants: oren rosenblatt,...
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![Page 1: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/1.jpg)
Ben Gurion University of the Negevwww.bgu.ac.il/atomchip
Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar
Week 4. Potential, capacitance and capacitors – E from V • equipotential surfaces • E in and on a conductor • capacitors and capacitance • capacitors in series and in parallelSource: Halliday, Resnick and Krane, 5th Edition, Chaps. 28, 30.
Physics 2B for Materials and Structural EngineeringPhysics 2B for Materials and Structural Engineering
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With 100,000 V on her body, why is this girl smiling???
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E from V
Let U(r2) be the potential energy of a charge q at the point r2. We found that it is minus the work done by the electric force Fq in bringing the charge to r2:
If we divide both sides by q, we get (on the left side) potential instead of potential energy, and (on the right side) the electric field instead of the electric force:
. )()()(2
1
12 r
r
rrFrr dUU q
. )()()(2
1
12 r
r
rrErr dVV
![Page 4: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/4.jpg)
E from V
Let V(r2) be the electric potential at the point r2. We have just obtained it from the electric field E:
Now let’s see how to obtain E from V. For r2 close to r1 we
can write r2 = r1 + Δr and expand V(r2) in a Taylor series:
. )()()(2
1
12 r
r
rrErr dVV
; ...)(
)()(
111
1
12
zz
Vy
y
Vx
x
VV
VV
zyxr
rrr
![Page 5: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/5.jpg)
E from V
We can write this expansion more compactly using the “del” (gradient) operator:
so
with all the derivatives evaluated at the point r1. We also write
, ,,
zyx
, ,,)( 1
z
V
y
V
x
VV r
. z)( 1
z
Vy
y
Vx
x
VV rr
![Page 6: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/6.jpg)
E from V
So now we can write
and therefore
...)()(
...z)()(
11
11
rrr
rrr
VV
z
Vy
y
Vx
x
VVV
. )()(
)()()(
)()()(
11
111
12
2
1
rrErr
rrErrr
rrErr
r
r
V
VV
dVV
![Page 7: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/7.jpg)
E from V
The equation
is a scalar equation, but since we can vary each component of Δr independently, it actually yields three equations:
)()( 11 rrErr V
, )()(
, )()(
, )()(
11
11
11
rr
rr
rr
zz
yy
xx
EV
EV
EV
![Page 8: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/8.jpg)
E from V
The equation
is a scalar equation, but since we can vary each component of Δr independently, it actually yields three equations, i.e.
)()( 11 rrErr V
, )()(
, )()(
, )()(
11
11
11
rr
rr
rr
z
y
x
EVz
EVy
EVx
![Page 9: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/9.jpg)
E from V
which reduce to a single vector equation:
, )()(
, )()(
, )()(
11
11
11
rr
rr
rr
z
y
x
EVz
EVy
EVx
. )()( rrE V
![Page 10: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/10.jpg)
E from V
Example 1 (Halliday, Resnick and Krane, 5th Edition, Chap. 28, Exercise 34): Rutherford discovered, 99 years ago, that an atom has a positive nucleus with a radius about 105 times smaller than the radius R of the atom. He modeled the electric potential inside the atom (r < R) as follows:
where Z is the atomic number (number of protons). What is the corresponding electric field?
, 22
31
4)(
2
2
0
R
r
Rr
ZerV
![Page 11: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/11.jpg)
E from V
Answer:
since
, ˆ1
4
1
4
22
31
4)()(
220
220
2
2
0
r
rE
R
r
r
Ze
rR
r
r
Ze
R
r
Rr
ZerV
. ˆˆˆˆ
ˆˆˆ rrzyx
zyx
rr
zyx
z
r
y
r
x
rr
![Page 12: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/12.jpg)
E from V
Example 2: Electric field of a dipole
We found that the electric potential V(x,y,z) of a dipole made of charges q at (0,0,d/2) and –q at (0,0,–d/2) is
z
–d/2
d/2
(x,y,z)
. )2
( re whe
, 1
1
4
),,(
222
0
dzyxr
rr
qzyxV
r+
r–
![Page 13: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/13.jpg)
E from V
. )2
( re whe
, 2
2
8
)(
, 1
1
4
)(
, 1
1
4
)(
222
330
330
330
dzyxr
r
dz
r
dzq
z
VE
rr
qy
y
VE
rr
qx
x
VE
z
y
x
r
r
r
z
–d/2
d/2
(x,y,z)
r+
r–
![Page 14: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/14.jpg)
E from V
Now we will consider the case d << r– , r+ and use these rules for 0 < α << 1 (derived from Taylor or binomial expansions) to approximate E:
. ...1) (1
, ... 1 1
1
, ... 2
1 1
nn
![Page 15: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/15.jpg)
E from V
For d << r– , r+ we have
z
(x,y,z)
r+
r–
. where, 4
1
4
2
2221/2
2
2
2
2222
222
zyxrr
d
r
zdr
dzdzyx
dzyxr
d/2
–d/2
![Page 16: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/16.jpg)
E from V
For d << r– , r+ we have
z
(x,y,z)
r+
r–
. 8
3
2
31
1
2
31
11
1
)(
1
so , 4
where, 1
2
2
2333/2
33
2
2
21/2
r
d
r
zd
rrrr
r
d
r
zdrr
d/2
–d/2
![Page 17: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/17.jpg)
E from V
So
z
(x,y,z)
r–
now and ; 3
8
3
2
31
1
8
3
2
31
11
1
5
2
2
232
2
2333
r
zd
r
d
r
zd
rr
d
r
zd
rrr
d/2
–d/2
. 4
)3( ,
4
3 ,
4
35
0
22
50
50 r
qdrzE
r
yzqdE
r
xzqdE zyx
r+
![Page 18: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/18.jpg)
E from V
We can check that these results coincide with the results we obtained for the special cases x = 0 = y and z = 0:
z
(x,y,z)d/2
–d/2 r+
. 4
, 2 3
03
0 r
qdE
z
qdE zz
. 4
)3( ,
4
3 ,
4
35
0
22
50
50 r
qdrzE
r
yzqdE
r
xzqdE zyx
r–
![Page 19: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/19.jpg)
E from V
This may look complicated but it is easier than calculating Ex,
Ey and Ez directly!
z
(x,y,z)
r–d/2
–d/2 r+
. 4
)3( ,
4
3 ,
4
35
0
22
50
50 r
qdrzE
r
yzqdE
r
xzqdE zyx
![Page 20: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/20.jpg)
Equipotential surfaces
We have already seen equipotential surfaces in pictures of lines of force:
and in connection with potential energy:r2
r1
r2
r1
rF(r)
![Page 21: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/21.jpg)
Equipotential surfaces
All points on an equipotential surface are at the same electric potential.
Electric field lines and equipotential surfaces meet at right angles. Why? r2
r1
r2
r1
rF(r)
![Page 22: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/22.jpg)
Equipotential surfaces
The surface of a conductor at electrostatic equilibrium – when all the charges in the conductor are at rest – is an equipotential surface, even if the conductor is charged:
Small pieces of thread (in oil) align with the electric field due to two conductors, one pointed and one flat, carrying opposite charges.
[From Halliday, Resnick and Krane]
![Page 23: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/23.jpg)
y
V
x
Equipotential surfaces
We can also visualize the topography of the electric potential from the side (left) and from above (right) with equipotentials
as horizontal curves.
![Page 24: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/24.jpg)
Equipotential surfaces
Quick quiz: In three space dimensions, rank the potential differences V(A) – V(B), V(B) – V(C), V(C) – V(D) and V(D) – V(E).
9 V
7 V
8 V
6 V
A
B
CDE
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E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
Explanation: A conductor contains free charges (electrons) that move in response to an electric field; at electrostatic equilibrium, then, the electric field inside a conductor must vanish.
Example: An infinite conducting sheet ina constant electric field develops surface charges that cancel the electric field inside the conductor.
+
+
+
+
+
+
+
-
-
-
-
-
-
-
![Page 26: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/26.jpg)
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
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E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
Explanation: A charge inside the conductor would imply, via Gauss’s law, that the electric field could not be zero everywhere on a small surface enclosing it, contradicting 1.
![Page 28: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/28.jpg)
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be normal to the surface and equal to σ/ε0, where σ is the surface charge density.
![Page 29: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/29.jpg)
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be normal to the surface and equal to σ/ε0, where σ is the surface charge density.
Explanation: A component of E parallel to the surface would move free charges around. (Hence the surface of a conductor at electrostatic equilibrium is an equipotential surface, even if the conductor is charged.)
![Page 30: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/30.jpg)
E in and on a conductor
Let’s compare the electric fields due to two identical surface charge densities σ, one on a conductor with all the excess charge on one side (e.g. the outside of a charged sphere) and the other on a thin insulating sheet.
The figure shows a short “Gaussian can” straddling the thin charged sheet. If the can is short, we need to consider only the electric flux through the top and bottom.Gauss’s law gives 2EA = ФE = σA/ε0, so
E = σ/2ε0. But if there is flux only through
the top of the can, Gauss’s law gives EA = ФE = σA/ε0 and E = σ/ε0.
“Gaussian can”
![Page 31: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/31.jpg)
E in and on a conductor
Four rules about conductors at electrostatic equilibrium:
1. The electric field is zero everywhere inside a conductor.
2. Any net charge on an isolated conductor lies on its surface.
3. The electric field on the surface of a conductor must be normal to the surface and equal to σ/ε0, where σ is the surface charge density.
4. On an irregularly shaped conductor, the surface charge density is largest where the curvature of the surface is largest.
![Page 32: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/32.jpg)
E in and on a conductor
Explanation: If we integrate E·dr along any electric field line, starting from the conductor and ending at infinity, we must get the same result, because the conductor and infinity are both equipotentials. But E drops more quickly from a sharp point or edge than from a smooth surface – as we have seen, E drops as 1/r2 from a point charge, as 1/r near a charged line, and scarcely drops near a charged surface. The integral far from the conductor is similar for all electric field lines; near the conductor, if E·dr drops more quickly from a sharp point or edge, it must be that E starts out larger there. Then, since E is proportional to the surface charge σ, it must be that σ, too, is larger at a sharp point or edge of a conductor.
![Page 33: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/33.jpg)
Capacitors and capacitance
A capacitor is any pair of isolated conductors. We call the capacitor charged when one conductor has total charge q and the other has total charge –q. (But the capacitor is then actually neutral.)
Whatever the two conductors look like, the symbol for a capacitor is two parallel lines.
capacitor
battery
switch
![Page 34: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/34.jpg)
Capacitors and capacitance
Here is a circuit diagram with a battery to charge a capacitor, and a switch to open and close the circuit.
capacitor
battery
switch
![Page 35: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/35.jpg)
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV.
capacitor
battery
switch
![Page 36: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/36.jpg)
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV. (Note that q and ΔV both scale the same way as the electric field.)
![Page 37: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/37.jpg)
Capacitors and capacitance
The charge q on the capacitor (i.e. on one of the two conductors) is directly proportional to the potential difference ΔV across the battery terminals: q = C ΔV. (Note that q and ΔV both scale the same way as the electric field.)
The constant C is called the capacitance of the capacitor. The unit of capacitance is the farad F, which equals Coulombs per volt: F = C/V.
![Page 38: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/38.jpg)
Capacitors and capacitance
Example 1: Parallel-plate capacitor
If we can neglect fringing effects (that is, if we can take the area A of the conducting plates to be much larger than the distance d between the plates) then E = σ/ε0 where σ = q/A and
ΔV = Ed = qd/ε0A. By definition, q = C ΔV, hence C = ε0A/d
is the capacitance of an ideal parallel-plate capacitor.
![Page 39: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/39.jpg)
Capacitors and capacitance
Example 2: Cylindrical capacitor
Again, if we can neglect fringing effects (that is, if we can take the length L of the capacitor to be much larger than the inside radius b of the outer tube) then
. ln ln
2C
, )(ln 22
, , 2
1)(
0
00
0
ab
L
b/aL
q
r
dr
L
qV
b r a Lr
qrE
b
a
b
a
![Page 40: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/40.jpg)
Capacitors and capacitance
Example 3: Spherical capacitor
The diagram is unchanged, only E(r) is different:
.
4C
, 11
44
, , 4
1)(
0
02
0
20
ab
ab
ba
q
r
drqV
b r a r
qrE
b
a
b
a
![Page 41: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/41.jpg)
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances. Here are two capacitors in series:
capacitor
battery
switch
![Page 42: Ben Gurion University of the Negev Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shai Inbar Week 4. Potential,](https://reader036.vdocuments.net/reader036/viewer/2022062714/56649d445503460f94a20927/html5/thumbnails/42.jpg)
Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances. Here are two capacitors in series:
Since the potential across both capacitors is ΔV, we must have q1 = C1ΔV1 and q2 = C2ΔV2 where
ΔV1 + ΔV2 = ΔV. But the charge
on one capacitor comes from the other, hence q1 = q2 = q.
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Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances. Here are two capacitors in series:
Since the potential across both capacitors is ΔV, we must have q = C1ΔV1 and q = C2ΔV2 where
ΔV1 + ΔV2 = ΔV. If the effective
capacitance is Ceff, then we have
hence
, 21
21eff C
q
C
qVVV
C
q
. 111
21eff CCC
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Capacitors in series and in parallel
For n capacitors in series, the generalized rule is
. 1
... 111
21eff nCCCC
C1 C2 Cn
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Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances. Here are two capacitors in parallel:
capacitor
battery
switch
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Capacitors in series and in parallel
Combinations of capacitors have well-defined capacitances. Here are two capacitors in parallel:
Since the potential across each capacitor is still ΔV, the charge on the capacitors is q1 = C1ΔV and q2 = C2ΔV. If the effective
capacitance is Ceff, then we have
C1ΔV + C2ΔV = q1 + q2 = CeffΔV,
thus capacitances in parallel add:
C1 + C2 = Ceff .
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Capacitors in series and in parallel
For n capacitors in parallel, the generalized rule is
Ceff = C1 + C2 +…+ Cn .C1
C2
Cn
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Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C. What is the equivalent capacitance Ceff of each of these combinations?
A B
C D
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A B
C D
Ceff = 3C
Ceff = C/3
Ceff = 2C/3
Ceff = 3C/2
Halliday, Resnick and Krane, 5th Edition, Chap. 30, MC 9:
The capacitors have identical capacitance C. What is the equivalent capacitance Ceff of each of these combinations?
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Halliday, Resnick and Krane, 5th Edition, Chap. 30, Prob. 9:
Find the charge on each capacitor (a) with the switch open and(b) with the switch closed.
ΔV =12 V
C1 = 1 μF C2 = 2 μF
C3 = 3 μF C4 = 4 μF
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Answer (a): We have C1ΔV1 = q1 and C3ΔV3 = q1 since those
two capacitors are equally charged. Now ΔV1 + ΔV3 = 12 V
and so q1/C1 + q1/C3 = 12 V and we solve to get q1 = q3 = 9 μC.
In just the same way we obtain q2 = q4 = 16 μC.
ΔV =12 V
C1 = 1 μF C2 = 2 μF
C3 = 3 μF C4 = 4 μF
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Answer (b): We have C1ΔV12 = q1 and C2ΔV12 = q2 and in the
same way C3ΔV34 = q3 and C4ΔV34 = q4. Two more equations:
ΔV12 + ΔV34 = 12 V and q1 + q2 = q3 + q4. We solve these six
equations to obtain ΔV12 = 8.4 V and ΔV34 = 3.6 V, and charges
q1 = 8.4 μC, q2 = 16.8 μC, q3 =10.8 μC and q4 = 14.4 μC.
ΔV =12 V
C1 = 1 μF C2 = 2 μF
C3 = 3 μF C4 = 4 μF