Electrostatic Formula

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Transcript Electrostatic Formula

Electrostatic
Formula
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Vector sum – x components, y components, sum of x components, sum of
y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )
Electric Force

F=
1
q1q2
4 p eo r2
=
k q 1q 2
r2
k = 9.0x10 9 N m2 = 1
= -------1-----------C2
4 p eo 4 p 8.85x10-12 C2
N m2
Vector sum – x components, y components, sum of x components, sum of
y components, pythagorean theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field associated with point
charges

E=F
q
F = kqq
r2
therefore E = Kqq therefore E =kq
r2q
r2
Vector sum – x components, y components, sum of x
components, sum of y components, pythagorean
theorem, and inv tan of (sum of y )/(sum of x )
Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E

q
d

Electric Field Associated with
Parallel Plates

The electric field is uniform between plates
Work is required to move +q from the – to +
W=Fd W=qEd W=qKqd W=Kqq W = Kq W =V=Joules= Electric Potential

d2
d
q
d q
Coulomb

W=qEd W = Ed V = Ed V = E = Volts

q
d
meter

Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance









C=Q
V
C = KAeo
d
K= dielectric constant
A = area in m2
e0= 8.85x10-12 C2
N m2
d = distance between plates in m
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance




W = QV = U ( Potential energy stored )
For a capacitor
U = ½ QV since it is easier to add charge at first and
progressively gets more difficult
C = Q CV = Q therefore U =1/2 QV= ½ CV2
V
C = Q V = Q therefore U = ½ QV = ½ Q2
V
C
C
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance


Cparallel = C1+C2+C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 12x10-6F (3V) = 36x10-6C or 3.6x10-5Q
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 12x10-6F(3V)2 =
U=1.08x10-4 J
What is the charged stored on each of the capacitors?
CV=Q= 2x10-6F (3V) = 6x10-6C
CV=Q= 4x10-6F (3V) = 12x10-6C or 1.2x10-5Q
3V
12mF
CV=Q= 2x10-6F (3V) = 18x10-6C or 1.8x10-5Q
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11=1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11=1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11=1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11=1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF
Capacitance
1


=1+ 1 + 1
Cseries C1 C2 C3
2mF
4mF
6mF
What is the total charge stored in the system? C = Q
V
CV = Q = 1.09x10-6F (3V) = 3.27x10-6C
What is the energy stored in the system?
U=1/2 QV U = ½ CV2 U= ½ 3.27x10-6F(3V)2 =
U=1.47x10-5 J
3V
1/C = 1/2mF 1/4mF + 1/6mF
1/C = 11/12mF = 12mF/11= 1.09 mF