Conductors and Insulators Capacitance & Capacitors

Energy Stored in Capacitors

W04D2

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Announcements

Week 4 Prepset due Week 4 Friday 8:30 am PS 4 due Week 5 Tuesday at 9 pm in boxes outside 26-152 Sunday Tutoring 1-5 pm in 26-152

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Outline

Conductors and Insulators Conductors as Shields Capacitance & Capacitors Energy Stored in Capacitors

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Conductors and Insulators

Conductor: Charges are free to move Electrons weakly bound to atoms Example: metals Insulator: Charges are NOT free to move Electrons strongly bound to atoms Examples: plastic, paper, wood

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Conductors are Equipotential Surfaces

1) Conductors are equipotential objects 2) E perpendicular to surface

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Properties of Equipotentials

E field lines point from high to low potential

E field lines perpendicular to equipotentials E field has no component along equipotential The electrostatic force does zero work to move a charged particle along equipotential

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Conductors in Equilibrium

Conductor placed in external electric field: charges in the conductor will move until 1)  E = 0 inside conductor 2) E perpendicular to surface 3) Induced surface charge distribution on surface of conductor in general is non-uniform

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Induced Charges on Conductors

Non-zero charge placed inside a metal slab (conductor): charges move to surface (move as far apart as possible)

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The Charged Metal Slab Applet: http://web.mit.edu/viz/EM/visualizations/electrostatics/CapacitorsAndCondcutors/chargedmetalslab/chargedmetalslab.htm

CQ: Metal Spheres Connected by a Wire

Two charged conducting spheres 1 and 2 with radii r1 and r2 are connected by a very long thin wire. What is the ratio of the charges q1/q2 on the surfaces of the spheres? You may assume that the spheres are very far apart so that the charge distributions on the spheres are uniform. 1.  (r1 / r2)2 2.  (r2 / r1)2 3.  r1 / r2 4.  r2 / r1

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Hollow Conductors: Applet Charge placed outside spherical shell conductor induces charge separation on outer surface of conductor. Electric field is zero inside both conductor and region interior to conductor.

http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm

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Demonstration:

Faraday Cage D33

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=D%2033&show=0

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Charging by Induction: Place external charge on the outside of a neutral metallic box (conductor): induced charges move to the surface of the conductor.

Induced Charge Distribution in External Electric Field

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http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/chargebyinductionBox/chargebyinductionBox.htm

Group Problem: Electric Field on Surface of Conductor

Consider a conductor with a positive non-uniform charge density σ on the outer surface. Choose a very thin Gaussian surface, where one end-cap is just above the surface, and the other end-cap lies inside the material of the conductor, as shown in the figure.

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(i)  What is the direction of the electric field on the surface? (ii)  Determine an expression for the electric field on the outer

surface of the conductor in terms of the surface charge density.

Group Problem Solution: Electric Field on Surface of Conductor

1)  E perpendicular to surface 2) Excess charge on surface Apply Gauss’s Law

EsurfaceA = σ Aε0

⇒ Esurface =σε0

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CQ: Charge inside cavity of Conductor I

1.  is initially uniform and does not change when the charge is moved.

2.  is initially uniform but does become non-uniform when the charge is moved.

3.  is initially non-uniform but does not change when the charge is moved.

4.  is initially non-uniform but does change when the charge is moved.

A point charge +Q is placed inside a neutral, hollow, spherical conductor. As the charge is moved around inside, the surface charge density on the outside

spherical conducting shell

+ Q

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CQ: Charge inside cavity of Conductor II

1.  is zero and does not change 2.  is non-zero but does not change 3.  is zero when centered but changes 4.  is non-zero and changes

A point charge +Q is placed inside a neutral, hollow, spherical conductor. As the charge is moved around inside, the electric field outside

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spherical conducting shell

+ Q

Hollow Conductors: Applet Charge placed inside interior of spherical shell conductor. Electric field outside is field of point charge.

http://web.mit.edu/viz/EM/visualizations/electrostatics/ChargingByInduction/shielding/shielding.htm 17

CQ: Charge inside cavity of Conductor III

1.  Q. 2.  Q + q. 3.  q. 4.  Q - q. 5.  Zero.

A point charge +q is placed inside a hollow cavity of a conductor that carries a net charge +Q. What is the total charge on the outer surface of the conductor?

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Capacitors and Capacitance

Our first of 3 standard electronics devices (Capacitors, Resistors & Inductors)

Capacitance Consider an isolated conductor with total charge charge Q. The conductor is an equipotential surface, with a certain potential V, where V = 0 at infinity. The charge on the conductor Q is linearly proportional to the potential The constant of proportionality C is called the capacitance and is only depends on the size and shape of the conductor. SI Units: 1 coulomb/volt = 1 farad

Q = CV

Q

conductor at potential V,(V = 0 at infinity)

conductor with charge Q,

+Q Q

Capacitor: Two Conductors

Capacitor: Two isolated conductors with equal and opposite charges ± Q. Potential difference ΔV between the two conductors. The capacitance of the two conducting surfaces is C is always a positive constant that just depends on the size and shape of the conductors

C = Q / ΔV

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Parallel Plate Capacitor

Gauss’s Law: C depends only on geometric factors A and d

ΔV = −

!E ⋅d!s

bottom

top

∫ = Ed = QAε0

d C = Q

ΔV=ε0 Ad

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E = Q / Aε0

CQ: Changing Dimensions

A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the charge on the plates? 1.  Q increases. 2.  Q is the same. 3.  Q decreases.

!

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CQ: Changing Dimensions

A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V? 1.  V increases. 2.  V is the same. 3.  V decreases.

!

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Demonstration:

Changing Distance Between Circular Capacitor Plates E4

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%204&show=0

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Group Problem: Spherical Shells spherical shell with charge

spherical shell with charge

a

bQ

+Q

A spherical conductor of radius a carries a charge +Q. A second thin conducting spherical shell of radius b carries a charge –Q. Calculate the capacitance.

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Group Prob. Solution: Spherical Shells spherical shell with charge

spherical shell with charge

a

bQ

+Q

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ΔV = −

!E ⋅d!s

a

b

∫ = − − Q4πε0r

2 r̂ ⋅d r̂a

b

∫ = + Q4πε0

drr 2

a

b

∫ = Q4πε0

1a− 1

b⎛⎝⎜

⎞⎠⎟

!E = − Q

4πε0r2 r̂

Gauss’s Law:

C = QΔV

=4πε0

1a− 1

b⎛⎝⎜

⎞⎠⎟

Isolated Spherical Conductor

Capacitance is . Other equipotential surface (second conducting surface) is located at infinity. In previous calculation set .

4πε0a

C = QΔV

=4πε0

1a− 1

b⎛⎝⎜

⎞⎠⎟

=4πε0

1a− 1∞

⎛⎝⎜

⎞⎠⎟

= 4πε0a

b = ∞

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Capacitance of Human Being

Model human being as a spherical conductor of radius a ≅ 1 m:

aC 04πε= mF1085.8 120

−×=ε

C ! 100×10−12 F ! 100 pF

You can measure your capacitance in a demonstration outside the Edgerton Center on the 4th floor of the infinite corridor.

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Energy To Charge Capacitor

1. Capacitor starts uncharged. 2. Carry +dq from bottom to top.

Now top has charge q = +dq, bottom -dq 3. Repeat 4. Finish when top has charge q = +Q, bottom -Q

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Stored Energy in Charging Capacitor

At some point top plate has +q, bottom has –q Potential difference is V = q / C Change in stored energy done lifting another dq is dU = dq V

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Stored Energy in Charging Capacitor

So change in stored energy to move dq is:

Total energy to charge to Q dU = dqV = dq q

C= 1

Cq dq

U = dU∫ = 1

Cq dq

0

Q

∫ = 1C

Q2

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Energy Stored in Capacitor

C = Q

ΔV

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2VCVQ

CQU Δ=Δ==

Where is the energy stored???

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Energy Stored in Capacitor

uE =

εoE2

2

Parallel-plate capacitor: C =

εo Ad

and V = Ed

Energy stored in the E field!

U = 1

2CV 2 = 1

2εo Ad

Ed( )2=εoE2

2× ( Ad) = uE × (volume)

Energy density [J/m3]

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Demonstration: Charging Up a Capacitor

U = 1

2CV 2 = 1

2(1×10−4 F)(4×103 V)2 = 800 J

A 100 microfarad oil-filled capacitor is charged to 4 KV and discharged through a wire Stored Energy:

http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%206&show=0 35