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Unformatted text preview: A A E 5 90E 5. INTRODUCTION TO ELECTRODYNAMICS C h5 – 1 A A E 5 90E 5.1 ELECTROSTATICS C h5 – 2 A A E 5 90E Where do charged particles come from? What are mechanisms to generate charge particles? Ionization by collisions: E le c tro n , Positive ion or neutral atom, Radiation (photo ionization). Thermal ionization, Surface ionization, Highfrequency electric fields, Electron attachment. What type of interaction do we observe with charged particles? C h5 – 3 A A E 5 90E Definitions/Observations Observations: There are two kinds of electric charges: Positive & Negative. Like charges repel each other; unlike charges attract each other. Charge Mobility: Frictional electricity, Motion strongly depending on the nature of material/substance. Definitions: A “Point Charge” is defined as a charged object whose radius is very small compared to the distance between it and other charged objects of interest. The special case, where all source charges are STATIONARY, though the test charge may be moving, is called ELECTROSTATICS. Conductors: charges move easily. Insulators: charge experience NO or little movement. C h5 – 4 A A E 5 90E Coulomb’s Law French physicist Charles Augustin Coulomb (17361806) 1785, Coulomb conducted experiments with a torsion balance to explore the law of forces between electrical charges. Coulomb establish a relationship between force and distance, quantitatively. F! 1 r2 Later expended to capture relative size of charges: F! q1q2 r2 Coulomb’s Law describes the electric force between two pointlike electrically charged objects: ! F= 1 4! " 0 k # Q1Q2 r
2 12 Magnitudes of Electric Charges #ˆ r ! r r Unit Vector ˆ = ! r Dialectic Constant ! F= F = 1 4! " 0 k # Q1Q2 r122 Permitivity of Free Space
! 0 = 8.85 " 10
#12 2 F #12 C = 8.85 " 10 m Nm 2 NOTE: Coulomb’s Law holds only for Point Charges!
C h5 – 5 A A E 5 90E Definition of Electric Field
+ qTest Definition: Faraday invented the concept of force field lines. He defined that: the direction of force lines coincides with direction of electric field at that point, the density of force lines is proportional to the strength of the field. + QSource +
! F2
+ q2 _ qTest ! E1
! ! F2 = q2 ! E1 The “virtual force”, which charged particles experience, is called ELECTRIC FIELD. + Electric Field: ! !F E= q !N $ #C & "% NOTE: The word field has a special meaning in mathematical physics. A field is a physical quantity that has a value at every location in space. Its value can be a scalar or vector.
C h5 – 6 A A E 5 90E Electric Field
FTest = 1 4! " 0 k # Q S qT rS2T = qT # E S
+ QSource Point Charge Coulomb’s Law: Electric Field: Direction/Patterns ES = 1 4! " 0 k # QS r2 _ qTest Definition: Positive direction of electric field points away from the positive charge. q + ˆ r ! r ! E
Point of Interest + ! E= q #ˆ r 2 4! " 0 k r # 1 C h5 – 7 A A E 5 90E Electric Field What can we conclude from equation?
z
S o u r ce P o in t T e s t C h a r ge Principle of Superposition q + 1 q _ 3 Q + T qi + q2
S o u r ce C h a r ge s ! ri ! r P F ie ld P o in t y
S o u r ce C h a r ge s Force on Test Charge: !!!! F = F1 + F2 + F3 + … Coulomb’s Law: ! qQ 1 Fi = # i 2 T # ˆi r 4! " 0k riT n ! q 1 F= # QT $ 2i # ˆi r 4! " 0k i =1 riT Electric Field: n ! qi 1 E= r $ r2 #ˆi 4! " 0k i=1 iT x ! E is a function of position, because the E separation vector depends on location of field point P. ! E is a vector quantity, which will vary with position. Configuration of source charges determine electric field. !! Physically: E ((r ) is the force per unit charge E that a test charge would experience at location P.
C h5 – 8 A A E 5 90E Electric Field of a Dipole
O b s e r va tio n P o in t Electric Field along the x–Axis:
y –
s + x r ! E!
P = (r, 0, 0) ! Ex ! E+ Coulomb’s Law:
E+ = 1 4 ! "0 # ( r $ s)
1 2 q+ 2 E! = 1 4 " #0 $ ( r + s)
1 2 q! 2 Superpositioning: E x = E+ + E! = q 4 " #0 $ ( r!1s 2 )(
2 2s r r+1s 2 ) 2 Approximation:
r ! s ! Ex " q 4 # $0 " 2s % 3 ! E x = E x 1, 0, 0 r ( )
C h5 – 9 A A E 5 90E Electric Field of a Dipole
! E+
1 2 1 2 Electric Field along the y–Axis: y Separation Vector for P:
! r+ = 0, t, 0 ! ! r!
1 2 1 2 ( ) ( s, 0, 0 ) = ( ! = ( 0, t, 0 ) ! ( ! s, 0, 0 ) = ( s, t, 0 ) s, t, 0 ) " " ! r+ = 12 4 s + t2
12 4 ! Ex
! E! P (0, t, 0) ! ! r! = r+ = s + t2 Unit Vectors: ! !1s t 0 r+ ˆ+ = ! = 2 r 12 r+ s + t2
4 ( ) ! r ˆ! = !! = … r r! –
s + x Superpositioning: ! ! ! "Ey " = E+ + E! = 1 4 " #0 $ ( q
12 4 s +t 2 ) 32 $ % ! 1 s, t, 0 ! &2 ( )( 1 2 1 s, t, 0 ' = ( 4" # $ 0 ) ( q
12 4 s +t 2 ) 32 ( ! s, 0, 0 ) Approximation: For location (0, r, 0): " q s t ! s ! Ey " # &31 0 0 4 $ %0 t ( ) ! 1 qs Ey ! " %31 0 0 4 # $0 r ( ) Comparison with Ex at (r, 0, 0): ! 1! "Ey " = E x 2
C h5 – 1 0 A A E 5 90E Dipole in an Electric Field
Q: What is the force on the dipole in a uniform field? A: Net force is zero! !!! ! ! F = F+ + F" = q E " q E = 0 ! Q: What will happen with the dipole in this field? A: Rotation about center of mass die to torque. T o rq u e :
!! !! T = r+ ! F+ + r" ! F" = Uniform Electric Field
! r+ s + ! E+ !
! E! – ! r! ( )( )( 1 2 ! ! ! ! !! s ! q E + " 1 s ! "q E = qs ! E 2 )( ( )) Definition of Dipole Moment: ! ! p!qs Non–Uniform Electric Field s + !!! ! ! ! F = F+ + F" = q E + " q E " = q #E $ 0 N e t F o rc e : ! ! !E represents the difference between the field at the plus end and the field at the minus end! Assumption: Dipole is very short s ! 1 – ! dE x = !E x " s ! dE y = !E y " s ! dE z = !E z " s ()
( ( ) ! " " "E # dE ) ! ! ! ! ! ! ! ! ! #E = s " ! E $ F = q #E = q s " ! E = p " ! E ( ) ( ) ( ) C h5 – 1 1 A A E 5 90E Electric Field of Distributed Charges Let’s consider a thin rod of length L and a total positive charge Q uniformly distributed on its surface: Q !Q = !y L Electric Field of a Line Charge
y ∆Q (0, y, 0)
! R Step I:
P (x, y0, 0)
! r
! !E ! r! Apply superposition principle. Assumption: rod thickness is very small ∆Q is small enough to be modeled as point charge! Step II: Choose origin, convenient location at half point. Derive expression for electric field due to one ∆Q.
• Separation Vector • Magnitude • Unit Vector • Electric Field
!!! R = r ! r " = ( x, y0 , 0 ) ! (0, y, 0 ) = ( x, y0 ! y, 0 ) (0, 0, 0) x ! R = x 2 + ( y0 ! y ) 2 = r ! R1 ˆ = ! = ( x, y0 ! y, 0 ) r Rr ! !E = 1 !Q $ ˆ = ( !E x , !E y , !E z ) r 4"# 0 r 2
C h5 – 1 2 A A E 5 90E Electric Field of Distributed Charges Step III: Problem simplification by considering symmetry Origin of problem is at (x,0,0): y0=0 S y m m e try Summation Electric Field of a Line Charge +y 2 E x = " !E x = "
! R2
! !E y1
P =(x, 0, 0) 1Q % 4#$ 0 L
+ L 2 x (x
2 2 +y 3 22 ) % !y = 1Q % x%" 4#$ 0 L !y (x 2 +y 3 22 ) ! !E1 ! !E
! !E y2 Ex = 1Q #x# 4!" 0 L L $ 2 % 1 (x +y 3 22 ) dy = 1 Q # 4! " 0 x # x 2 + (L 2 )2 ! !E 2 ! R1
–y
1 ! E= 1 Q # #ˆ r 4! " 0 r # r 2 + (L 2 )2 Step IV: Case Study C a se I: C a se II: r >> L L >> r ! E= ! E= 1Q #ˆ r 4!" 0 r 2 1 2Q L ˆ # r 4!" 0 r
C h5 – 1 3 A A E 5 90E Electric Field of Distributed Charges Electric Field of a Thin Ring
z P (0, 0, z) See Homework #6 Solutions! x ∆Q
y
C h5 – 1 4 A A E 5 90E Electric Field of Distributed Charges Electric Field of a Charged Disk
z P (0, 0, z) See Homework #6 Solutions! x ∆Q y
C h5 – 1 5 A A E 5 90E Electric Field of Distributed Charges Electric Field of Uniformly Charged Spherical Shell ! r + + +
R + + + See Homework #6 Solutions! + + C h5 – 1 6 A A E 5 90E Distributed Charges  General
!! E( r ) = 1 r # r2 ˆ $ dq 4! " 0 ! #(r$) r & r2 ˆ % dl 4! " 0 P 1 ! # (r$) r & r2 ˆ % da 4! " 0 S 1 ! 1 #( r $ ) r ' r2 ˆ % d& 4! " 0 V 1 General Line Charge !! E( r ) = dq = # % dl #= Q L Surface Charge !! E( r ) = dq = # % da #= Q A Volume Charge !! E( r ) = dq = # % d& #= Q V C h5 – 1 7 A A E 5 90E Distributed Charges  Special Cases
! E= 1 4! " 0 # Q r # r 2 + L2 4 z Thin Rod ( ) 1 2 #ˆ r Assumption: Symmetry r!L Case I: Case II: r !L Thin Ring ! Q E= # 4! " 0 (R 2 +z 3 22 ) #ˆ r Along symmetry axis for a ring of radius R. z=0 Case I: z!0 Case II: Case III: z=R Disk % ! 1 Q' z E= # 2 # '1 $ 2! " 0 R ' R 2 + z2 & ( ) ( * #ˆ r 1* 2* ) Along symmetry axis for a disk of radius R. z=0 Case I: 0 < z !R Case II: z R !1 Case III: Sphere ! E= 1 4! " 0 ! E=0 # Q #ˆ r 2 r r >R r <R Collection of Charges ! Ei = $r 4! " k
0 i =1 1 n qi
2 i # ˆi r
C h5 – 1 8 A A E 5 90E Patterns of Field in Space Examination of different charge distributions: Collection of point charges, line, surface and volume charges, Developed tools and equations to calculate the electric field due to those charges, Collectively these charge distributions produce a certain electric field. Coulomb’s Law is the governing law in electrostatics. Let’s consider the following cases: There is a connection between the pattern of electric field on a closed surface and the amount of charge inside that surface.
C h5 – 1 9 A A E 5 90E (Electric) Flux Definition of Flux: !!! ˆ ! = E " #A = E " n #A Properties of Flux: Direction of Electric Field Positive: electric field points out, Negative: electric field points in. Magnitude of Electric Field Surface Area C h5 – 2 0 A A E 5 90E Gauss’s Law German mathematician and physicist Carl Friedrich Gauss (17771855) Gaussian Surface can take any shape, Most useful surface mimics the symmetry of the problem at hand, Surface must be closed!! Definition: ! E
Surface of a Sphere Gaussian Surface ! ! ! = $ E i " #A i !! ! = " E " da %
S i r q + NOTE: The electric flux through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. Point Charge: !! 1q ˆ ! r 2 sin % d% d& ˆ E ! da = " r r " 2 4#$ 0 r S ( ) !!q " E ! da = = $
S (infinitesimal) surface of sphere #0 C h5 – 2 1 A A E 5 90E Gauss’s Law Gauss’ Law relates the net flux Φ of an electric field through a closed surface to the net charge Qenc that is enclosed by that surface. Gauss’ Law in Integral Form: ! ! Q enc ! = " E " da = # $0 S NOTE: Gauss’s Law does not depend on the surface size or shape. Applying Fundamental Theorem for Divergence: Gauss’s Theorem !! ! % E ! da = " # ! E d$ = " d$ "
S V ( ) V &0 where charge density: != Q V or Q = # ! d"
V Gauss’s Law in Differential Form: !# !"E = $0
C h5 – 2 2 A A E 5 90E Class Example #2  Gauss’ Law ! A nonuniform electric field given by E = 3.0 x ˆ + 4.0 ˆ pierces the Gaussian i j cube shown in figure below. What is the electric flux through the right face, the left face and the top face? What is the net flux?
A TFˆ j A RFˆ i A FF ˆ k C h5 – 2 3 A A E 5 90E Summary Electric Force acts on Point Charges: Coulomb’s Law, Electric Charges make Electric Fields, Electric Fields depend on Charge Distribution, Electric Fields (magnitude and direction) are strong functions of Position, Knowing Electric Field Distribution over a surface allows us to determine quantity of Charge and in certain cases Charge Distribution: Gauss’ Law C h5 – 2 4 A A E 5 90E Curl of Electric Field
! E
Q Electric Field of a Point Charge: ! 1q E= # 2 #ˆ r 4! " 0 r + Could we construct/place charges in such a way to produce an electric field, which would look like this? Can we come up with an example/method to prove it to us a little more rigorously? Can we show that ! !"E =0
C h5 – 2 5 A A E 5 90E Curl of Electric Field
y Let’s consider the electric field of a point charge and calculate the line integral of this field from some point a to b:
!! " E ! dl
b a b +Q
z +
a x Electric Field by a Point Charge: ! E= 1 4! " 0 # q #ˆ r 2 r Spherical Coordinates: ! ˆ ˆ d l = dr ˆ + r d! ! + r sin ! d" " r
q 1 dr = 4# $ 0 % r 2 4# $ 0 a 1
b ! !! E " dl = q " 2 dr ! 4# $ 0 r 1 ' q *b 1qq & ,= (&) )r 4# $ 0 ra rb ( + ra r C h5 – 2 6 A A E 5 90E Curl of Electric Field What will happen if we complete a closed loop integral and return to our starting point?
!! " E ! dl =
P qq ( % )=0 4# $ 0 ra ra 1 How does this result help us with the curl of an electric field? Stokes’ Theorem: #(
S !! ! !! ! " E $ da = " E $ d l = 0 # ) ! ! "#E =0
+Q
z P y b +
a x C h5 – 2 7 A A E 5 90E Energy in Electric Fields Force: Potential Energy: FC = 1 4! " 0 # Electric qQ r2 (Coulomb’s Law) FN = G ! Gravitational mM r2 (Newton’s Law) Remember: Potential Energy is energy that can be associated with the configuration/arrangement of a system of objects that exert forces on one another!
Electric Potential Energy is associated with the state of separation between charged particles, which attract/repel one another via the electric force. Gravitational Potential Energy is associated with the state of separation between objects, which attract one another via the gravitational force. U el = 1 4! " 0 # qQ = qE r r U gr (h ) = m g h Work: System changes configuration from an initial state i to a different final state f, the force in the system does work W on the object/particle.
!U = U f " U i = "W
C h5 – 2 8 A A E 5 90E Electric Potential
! E
+Q
q1 q2=2q1
1st Observation Point 2nd Observation Point Potential Energy at: q1 q2=2q1 1st Observation Point: (U (U 2 = 2U1 ) ) 1OP 2nd Observation Point: 2 = 2U1 2 OP ⇒ We find that the potential energy per charge at our observation point has the same value independent of the charge size and is characteristic only of the electric field we are interested in. C h5 – 2 9 A A E 5 90E Electric Potential
! E
+Q
q1 q 2 = 2q 1 1 s t O b se r va tio n P o in t 2 n d O b s e r va tio n P o in t q1 q 2 = 2q 1 Definition of Electric Potential: U V= q ! J$ #V = C & " % NOTE: Electric Potential is a SCALAR, not a vector! Electric potential difference ΔV between any two points (initial i and final f) in an electric field is: !U W !V = Vf " Vi = =" q q Thus, the potential difference between two points is the negative work done by the electrostatic force to move a unit charge from one point to the other. Potential difference can be positive, negative, or zero, depending on the signs and magnitudes of q and W. C h5 – 3 0 A A E 5 90E ! Sign of ∆V and Direction of E You are holding a positive charge in your hand. Q: What happens when you simply release it? How does the positive charge move? Positive charge starts to move toward lower potential in the direction of electric field. Q: What happens if you push the positive charge against the electric field? You are moving it to higher potential, you invest work. Q: What happens when you simply release a negative charge? A negative charge that is free to move will accelerate in the opposite direction of the electric field. Therefore, a negative charge heads toward a region of higher potential. Definition: Moving in the same direction of E means that the potential is decreasing! Moving in opposite direction of E means that the potential is increasing! C h5 – 3 1 A A E 5 90E ! Examples: What is the direction of E and is ∆V positive or negative? ! Sign of ∆V and Direction of E C h5 – 3 2 A A E 5 90E ! Relationship between E and V E Definition: Constant potential: along such a surface the voltage does not change, Perpendicular to electric field, W=0 for any path between an initial and final point both located on the same equipotential surface. So far we have established: Electric field is “force per unit charge”: ! !F E= q Potential is “potential energy per unit charge”: U V = el q Moving in the direction of E, potential decreases! ! E is ! to equipotential surfaces at any point P.
C h5 – 3 3 A A E 5 90E Electric Potential Calculating Electric Potential from an Electric Field Work done on charged particle displaced from i to f: !! !! dW = F ! d s = q E ! d s f !! W = " q E ! ds Potential Difference between two points: f !! ! (Vf ! Vi )q = # q E " d s
i
i !! Vf ! Vi = ! # E " d s
f i ! E Choosing reference point Vi = 0: Resulting integral is independent of path! f !! V = ! # E " ds
i NOTE: Electric Potential and Electric Potential Energy Electric Potential is a Property of an electric field, regardless of whether a charged object has been places in that field. Electric Potential Energy is the energy of a charged object in an external electric field, or more precisely, energy of the system (object of interest and external elect. field). C h5 – 3 4 A A E 5 90E Electric Potential Calculating Electric Field from an Electric Potential ! Positive charge moves through displacement d s : Work done by electric field d
dW = ! q dV !! !! dW = F " d s = ( q E ) " d s = q E " cos# " ds
! q dV = q E " cos# " ds dV E " cos# = ! ds The electric field E in any direction is the negative gradient of the electric potential with distance in that direction! "V ES = ! "s Considering our result with regard to Cartesian coordinate system: "V "V "V Ex = ! Ey = ! Ez = ! "x "y "z !V ! ! E = "#V Extraordinary Observation: ! E is a vector quantity (three components), E V is a scalar quantity (one component)! C h5 – 3 5 A A E 5 90E Electric Potential How can one function possibly contain all the information that three ! independent functions carry? Further, are the three components of E really as independent as they look?
! !"E =0 The answer brings us back to the curl of the electric field:
! %# # ( %# # ( %# #( ! " E = ' Ez $ Ey * $ ' Ez $ E x * + ' Ey $ E x * = 0 #z ) & #x #z ) & #x #y ) & #y + #E x #E y = #y #x #E z #y = #E y #z #E x #E z = #z #x E is a very special kind of vector! Potential formulation reduces a vector problem down to a scalar problem. C h5 – 3 6 A A E 5 90E ! Fundamental Equations for E
! !F E= q
! E = !"V Electric Field: Gradient: Divergence: Gauss’ Law Curl: !# !"E = ! !"E =0 $0 We used the curl law to show that E could be expressed as the gradient of a scalar: ! ! !"E = 0 ! " E = 0 permits ! ! E = !"V guarantees E = !"V Poisson’s Equation: Laplace’s Equation: !2 V = " # $0 !2 V = 0 Laplace’s Equation is a special case of Poisson’s Equation! Region where there is no charge: ρ = 0.
C h5 – 3 7 A A E 5 90E Class Example #3 A solid nonconducting sphere of radius R has a nonuniform charge distribution of volume charge density ! = !s r R , where ρs is a constant and r is the distance from the center of the s p h e re . a.) Show that the total charge on the sphere is Q = ! " sR . 3 b.) Show that the magnitude of the electric field inside the sphere is E = Q2 r 4! " 0 R 4 # 1 !( r ) = !s "r R !( r ) = !s "r R Gaussian Surface R R C h5 – 3 8 A A E 5 90E Work and Energy
Collection of Source Charges Work done to move a charge:
Q: How much work will you have to do? ! ! Electric Force on the test charge q0: Fel = q0 E ! Force you must exert is opposite the electric force: ! q0 E Work done by you: f f !! !! W = " Fel ! d s = # q0 " E ! d s = q0 Vf # Vi
i i f i ( ) Work you have to do to bring a charge q0 from “infinity” (reference point) to a ! point at r: r ! r3 ! ! ! r2 W = q0 Vf ! Vi = q0 V ( r ) ! V ( " ) = q0V ( r ) ( ) ( ) Work to assemble a collection of point charges: ! r1 ! W2 = q2V ( r ) = q1 q 4! " 0 2 r12 1
#q q& q 1 W4 = q4 % 1 + 2 + 3 ( 4! " 0 $ r14 r24 r34 ' q3
r23 r13 r12 q1 #q q& 1 W3 = q3 % 1 + 2 ( 4! " 0 $ r13 r23 ' q2 % ( 1 n ' n 1 qj * 1 n ! W = $ qi $ = $ q iV ( ri ) 2 i=1 ' j=1 4!" 0 rij * 2 i=1 & j# i )
C h5 – 3 9 A A E 5 90E Work and Energy
1 !V d" 2#
$0 Energy of a Continuous Charge Distribution: W= W=
W=
W= 1 $ V da 2#
!0 W= 1 %V dl 2# !# !"E = 2 %( ! ! !! !0 ' & E # "V d$ + VE # da ) " # E V d$ = ( % " % * 2 ) ! !V = " E !0 % ! !( E 2 d" + " V E $ da * '# # 2 &V S )
E A sphere V Total Energy stored in a Charge Configuration: 1" $ r2 $ ! !* 2$ ! r # lim , " V E ( da / = 0 ) r& ' +S . $ 1$ ! $ r% ! W= !0 2 all space # E 2 d"
! 0 Energy stored in the field with at density: E 2 = energy per unit volume 2 Because electrostatic energy is quadratic in the field, it does not obey superposition principle. C h5 – 4 0 A A E 5 90E Example  Parallel Plates
+Q
! E+ Energy Density of Electric Field between two Parallel Plates (Capacitor): Consider moving one plate of a capacitor. Electric Field due to positive plate: Force one plate exerts on the other: –Q ! QA E+ = 2! 0 ! ! QA F! = Q ! E + = Q " 2# 0 ! Fon "! " plate
by " + " plate Work you invest to pull negative plate from the positive increases potential energy: !U = W = F " !s = Q "
2 QA " !s 2# 0 1 $Q A' 1 !U = # 0 & " A !s = # 0E 2 !Volume 2 % #0 ) 2 ( Energy per Unit Volume:
!U 1 = " 0E 2 !Volume 2 C h5 – 4 1 A A E 5 90E Conductors Definition: One or more electrons per atom (of material) are free to move through the material. A perfect conductor contains an unlimited supply of completely free charges. Electrostatic Properties of Ideal Conductors: E=0 inside a conductor. ρ=0 inside a conductor. Any net charge resides on surface. A conductor is an equipotential. E is ⊥ to surface just outside a conductor. C h5 – 4 2 A A E 5 90E Capacitor
+Q –Q Definition: Two isolated conductors of any shape are basic elements of a capacitor. Potential Difference between the Surfaces (taking into account properties of conductors): Electric Field (arbitrary shape) given by Coulomb’s Law: !! V = V+ ! V! = ! # E " d l
+ ! V $E ! E &Q
V !Q ! E= 1 4! " 0
Q V $r #
2 ! r d%
" C% F= ' $ V& # Capacitance: C! Capacitance expresses the proportionality constant of the arrangement. Capacitance is purely a geometry quantity determined by shape, size, separation. C h5 – 4 3 A A E 5 90E Capacitor
q = ! 0E A
+ Capacitance of a ParallelPlate Capacitor Gauss’ Law: Potential: Capacitance: V = " E ds = E " ds = E d d E= C= !0 A ! 0 V d d Capacitance of a Cylindrical Capacitor Gauss’ Law: Potential: Capacitance: q = ! 0E A = ! 0E " 2# r L + b % b( q 1 q V = " E ds = ! dr = ln 2# $ 0L " r 2# $ 0L ' a * &) ! a L C = 2! " 0 ln b a () Capacitance of a Spherical Capacitor Gauss’ Law: Potential: Capacitance: q = ! 0E A = ! 0E " 4# r 2 + b q 1 q b!a V = " E ds = ! dr = % 4# $ 0 " r 2 2# $ 0 ab ! a ab C = 4! " 0 b#a C h5 – 4 4 ...
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