Charging capacitors Storing energy and charging

Book page 458, 463 - 470

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Review • Series

• 𝑉𝑡 = 𝑉1 + 𝑉2 + 𝑉3

• Q is the same

•1

𝐶𝑒𝑞𝑢=

1

𝐶1+

1

𝐶2+

1

𝐶3

• 𝑅𝑒𝑞𝑢 = 𝑅1 + 𝑅2 + ⋯

• Parallel

• 𝑉𝑡 = 𝑉1 = 𝑉2 = 𝑉3

• 𝑄𝑡 = 𝑄1 + 𝑄2 + 𝑄3

• 𝐶𝑒𝑞𝑢 = 𝐶1 + 𝐶2 + ⋯

•1

𝑅𝑒𝑞𝑢=

1

𝑅1+

1

𝑅2+

1

𝑅3

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𝐶 =𝑞

𝑉

𝐶 = 𝑘𝜀0

𝐴

𝑑

Where k = dielectric constant = 𝜀

𝜀0

𝜀0 = 𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒 = 8.9 × 10−12

Energy changes

• From experimental data a graph of charge q stored on a capacitor vs PD can be drawn

• Q = It

• The result is a straight line with the gradient equal to C

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Definition

• PD is the energy per unit charge stored on the plates: 𝑃𝐷 =𝐸

𝑞

• QV = energy stored = Area under the graph

= 1

2× 𝑞 × 𝑉

• QV = 1

2× 𝑞 ×

𝑞

𝐶

• Energy stored = 1

2

𝑞2

𝐶

• Energy stored can be written as 1

2𝐶𝑉2

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But Q = CV

Example

• A 6V battery is connected to a 20𝜇𝐹 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟. Find the energy that can be stored on the capacitor.

• Solution

• E = 1

2𝐶𝑉2

=1

2× 20 × 10−6 × 62

= 3.6 × 10−4𝐽

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Resistor – capacitor RC series circuits • In a basic resistor circuit, charge is immediately available to be used

by the resistor and it has a constant value

• 𝐼 =𝑒𝑚𝑓

𝑅

• 𝑄 = 𝐶𝑉

• 𝐼 =𝑄

𝑡=

𝑒𝑚𝑓

𝑅

• In a RC circuit charge is building up over time

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What happens in a charging RC circuit

t = 0 𝒕 → ∞ 𝒇𝒖𝒍𝒍𝒚 𝒄𝒉𝒂𝒓𝒈𝒆𝒅

∆𝒕 during charging

𝐼0 =𝑒𝑚𝑓

𝑅

I = 0 𝐼 =

∆𝑄

∆𝑡

𝑉𝑅 = 𝑒𝑚𝑓 𝑉𝑅 = 0

𝑉𝐶 = 0 𝑉𝐶 = 𝑒𝑚𝑓

𝑄𝐶 = 0 𝑄𝐶 = 𝑄𝑓𝑖𝑛𝑎𝑙 = 𝐶 × 𝑒𝑚𝑓

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• Capacitor starts charging, battery starts to charge the plates • Charge passes through resistor • Circuit carries a changing current • At max value capacitor has charge Q = C x emf • Current in circuit will be zero

Max charge on capacitor

• Maximum charge on capacitor = capacitance x emf of cell

• The current does not flow through the capacitor

• Why?

• It looks like it is flowing

• During charging: 𝐼𝑐 ↓

𝑉𝐶 ↑ until max emf

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What happens at the plates • An electric field is set up between the plates

• But: no electrons flow through the material

Simplify

• 1. electric current – flow of charge and electrons

• 2. displacement current (Maxwell) flows through capacitor new definition

• Charging a capacitor changes the electric field b/w the plates

• It also creates a changing magnetic field with enough energy to move electrons

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Charging a capacitor

• Current puts positive charge on capacitor, Q = CV

• Positive charge exerts force that pulls the negative charge

• They babance out

• Galvanometer shows direction

• Electric field E = 0 when start charging and slowly increases in time

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Charging continued

• V = Ed

• C = 𝜀𝐴

𝑑

• I = C ×𝑑𝑉

𝑑𝑡= 𝐽𝐴 where J – current density and A = area

• 𝐽𝐴 = 𝜀𝐴

𝑑×

𝑑

𝑑𝑡𝐸𝑑

• 𝐽 = 𝜀 ×𝑑𝐸

𝑑𝑡=

𝑑𝐷

𝑑𝑡, where

𝑑𝐷

𝑑𝑡 = displacement current density,

the rate of change of electric flux vector with time

• This way the current is related to the change of a field ©cg

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Re-conceptionalize current

• E puts force on molecules

• Negatively charged electron clouds move to positive plate

• Shifting of electron clouds act like a current

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Charging plates

• Fill the space with dielectric material

• It contains atoms and molecules

• Even in vacuum an electric flux vector will be set up, which acts exactly like a current is flowing

• This is the displacement current

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Graphical interpretation of charging

• 𝐼0 = gradient of graph at origin

• As charge build up, current drops

• At maximum charge, I = 0

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Time constant

• In principle, a capacitor can never fully charge because the rate of charging drops as the charge increases

• In practice, after a finite time the charging current becomes too small to be measured and we say the capacitor is fully charged

Definition:

• The time it takes to charge a capacitor in a given circuit is determined by the time constant 𝜏 of the circuit

• 𝜏 = 𝑅𝐶

• The half time 𝑡1

2

is the time taken to halve the charging

current

• 𝑡1

2

= 𝜏 × ln 2 = 𝑅𝐶 ln 2 = 0.693𝑅𝐶

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Example • A 250𝜇𝐹 capacitor is being charged from a 6V cell of negligible

internal resistance via a 10kΩ resistor. What is the initial charging current?

• Solution

• 𝐼 =𝑒𝑚𝑓

𝑅=

6

10000= 0.6 × 10−3𝐴 = 0.6𝑚𝐴

• What is the charging current after 10.4s?

• Solution

• 𝜏 = 𝑅𝐶=250 × 10−6 × 10000 = 2.5𝑠

• 𝑡1

2

= 𝜏 × ln 2 = 𝑅𝐶 ln 2 = 0.693𝑅𝐶 = 0.693 × 2.5 = 1.73𝑠

• 𝑛 =𝑡

𝑡12

=10.4

1.73= 5.78~6 ℎ𝑎𝑙𝑓 𝑙𝑖𝑣𝑒𝑠

• Charging current after 6 𝑡1

2

=𝐼0

2𝑛 =0.6×10−3

26 = 9.4𝜇𝐴

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𝐼 =𝐼0

2𝑛

Mathematics of charging a capacitor

• Kirchhoff’s 2nd law: Emf = V + IR

• V and I are changing charge on capacitor:

Q = CV or 𝑉 =𝑄

𝐶

• 𝐼 =𝑑𝑄

𝑑𝑡

• This is positive because charge on plates increases with time

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+q

- q

Putting it together • Emf= 𝑉 + 𝐼𝑅 =

𝑄

𝐶+ 𝑅

𝑑𝑄

𝑑𝑡

•𝑑𝑄

𝑑𝑡=

𝑒𝑚𝑓

𝑅−

𝑄

𝑅𝐶=

𝑒𝑚𝑓𝐶−𝑄

𝐶𝑅

• Let U = 𝑒𝑚𝑓𝐶 − 𝑄, then emf and C are constants

•𝑑𝑈

𝑑𝑡= −

𝑑𝑄

𝑑𝑡 where

𝑑𝑄

𝑑𝑡=

𝑈

𝐶𝑅

•𝑑𝑈

𝑑𝑡=

𝑈

𝐶𝑅

• Integrating this expression we get

• 𝑈 = 𝑈0𝑒−𝑡

𝐶𝑅

• 𝑈0= value when t = 0

• When t = 0 Q = 0

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• When t = 0 we get

• 𝑈 = 𝑈0𝑒−𝑡

𝐶𝑅

• emf C – Q =(emfC-0) 𝑒−𝑡

𝐶𝑅

• Q=emfC - emfC 𝑒−𝑡

𝐶𝑅

• Q = emf C(1 - 𝑒−𝑡

𝐶𝑅) • Q = emf C

• Q = 𝑄𝑓𝑖𝑛𝑎𝑙 1 − 𝑒−𝑡

𝐶𝑅

• CR = 𝜏

• Q = 𝑄𝑓𝑖𝑛𝑎𝑙 1 − 𝑒−𝑡

𝜏

Mathematical skills

• Q = 𝑄𝑓𝑖𝑛𝑎𝑙 1 − 𝑒−𝑡

𝜏

•𝑄

𝑄𝑓𝑖𝑛𝑎𝑙= 1 − 𝑒−

𝑡

𝑅𝐶

• 1 - 𝑄

𝑄𝑓𝑖𝑛𝑎𝑙= 𝑒−

𝑡

𝑅𝐶

• ln 1 − 𝑄

𝑄𝑓𝑖𝑛𝑎𝑙=

ln 𝑒−𝑡

𝑅𝐶 = −𝑡

𝑅𝐶

• 𝑡 = −𝑅𝐶 ln 1 − 𝑄

𝑄𝑓𝑖𝑛𝑎𝑙

• Suppose RC = 5.00s

Q = 1

4 𝑄𝑓𝑖𝑛𝑎𝑙

• The time required for the capacitor to acquire this charge is

• 𝑡 = −𝑅𝐶 ln 1 − 𝑄

𝑄𝑓𝑖𝑛𝑎𝑙

= −5[ln(1 −1

4)] = −5ln

3

4

= -5 x (-0.288) = 1.44s

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Example

• The time constant for (A) is 0.20s. What is the time constant for (B)?

• Solution

• Find 𝐶𝑒𝑞𝑢 for each arrangement

• A) 1

𝐶𝑒𝑞𝑢=

1

𝐶+

1

𝐶=

2

𝐶

𝐶𝑒𝑞𝑢 =1

2𝐶

• B) 𝐶𝑒𝑞𝑢 = 𝐶 + 𝐶 = 2𝐶 • Time constant 𝜏𝐴 = 0.20

• 𝜏 = 𝐶𝑅 → 0.20 =1

2𝐶R

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A

R C C

B

• 𝜏𝐵 = 2𝐶𝑅

•𝜏𝐵

𝜏𝐴=

2𝐶𝑅1

2𝐶R

• 𝜏𝐵 = 4 × 0.20 = 0.80𝑠

EXAMPLE: Two capacitors are in the circuit

shown. C1 is fully charged at 1.50 V, and C2

is initially uncharged.

(a) What is the electrical energy stored in C1?

(b) What is the charge on C1’s plates?

SOLUTION:

(a) E = 1

2CV 2 = 3.0910-4 J

(b) q = CV = 27510-6(1.50) = 4.12510-4 C.

Solving problems involving parallel-plate capacitors

Example

EXAMPLE: Two capacitors are in the circuit

shown. C1 is fully charged at 1.50 V, and C2

is initially uncharged.

(c) When the switch is closed, what will the

charge on C1 be? What charge will C2 have?

SOLUTION: The capacitors are in parallel. Thus

V1 = V2 q1

C1

= q2

C2

Thus q2 = C2

C1

×q1.

From conservation of charge, q1 + q2 = 4.12510-4 C.

4.12510-4 = q1 + q2 = q1 + C2

C1

×q1.

= q1 ( 1 + 38275

) q1 = 3.62 10-4 C.

3.62 10-4 + q2 = 4.12510-4 q2 = 5.01 10-5 C.

Example continued

PRACTICE: A 1.50-V cell is connected

to a 275 F capacitor in the circuit

shown.

(a) Make a sketch graph of what the

voltmeter reads after the switch is closed.

SOLUTION:

Before the switch is closed the

voltage on the capacitor is zero.

After the switch is closed, the

voltage will increase from zero to 1.50 V in a very short

amount of time as the plates store the charge.

Then the capacitor’s voltage will remain at 1.50 V.

Charging a capacitor

V / V

t

1.50

Example

PRACTICE: A 1.50-V cell is connected

to a 275 F capacitor in the circuit

shown.

(b) Make a sketch graph of what the

ammeter reads after the switch is closed.

SOLUTION:

Before the switch is closed the

current is zero.

After the switch is closed, the

current will begin at a maximum value and will decrease

rapidly as the plates “fill up” with charge and repel

further charge.

Then the current will remain at zero.

Charging a capacitor

I / A

t

MAX

Example continued

PRACTICE: A 1.50-V cell is connected

to a 275 F capacitor in the circuit

shown.

(c) Make a sketch graph of the charge q

on the plates after the switch is closed.

SOLUTION:

Before the switch is closed the

charge is zero.

After the switch is closed, the

charge will rapidly “fill up” the plates and slow down the

subsequent flow due to repulsion.

Then the charge will reach a maximum value on the

plates, stopping the flow of further charge.

Charging a capacitor

q / C

t

MAX

Example contiuned

PRACTICE: A 1.50-V cell is connected

in series to a capacitor and resistor.

Make a sketch graph showing the family

of curves representing the voltage across

the resistor after the switch is closed as RC increases.

SOLUTION:

Before the switch is closed the

current is zero.

After the switch is closed, the current will begin at a

maximum value and will decrease rapidly as the plates

“fill up” with charge and repel further charge.

Then the current will remain at zero.

RC series circuits – charging

I / A

t

MAX

Example

C

R

PRACTICE: A 1.50-V cell is connected

in series to a capacitor and resistor.

Make a sketch graph showing the family

of curves representing the voltage across

the resistor after the switch is closed as RC increases.

SOLUTION:

Because the voltage across

the resistor is proportional to

the current (V = IR), its graph will have the same shape

as I:

Furthermore, the bigger R and

C are, the longer it will take for

the capacitor to charge and the current to stop.

RC series circuits – charging

I / A

t

MAX

V / V

t

1.50

low RC

medium RC high RC

Example

C

R