Microsoft PowerPoint - Phys CH 23(96) []2323--1 What is Physics1 What is Physicsyy
One of the primary goals of physics is to find simple ways of solving seemingly complex problems One of the main toolssolving seemingly complex problems. One of the main tools of physics in attaining this goal is the use of symmetry.
For example in the ring and disk charged distributionsFor example, in the ring and disk charged distributions
0 0 ==== ∫∫ yyxx dEE,dEE yy
For certain charge distributions involving symmetry, we can save far more work by using the so-called Gauss’ law (
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save far more work by using the so-called Gauss law ( ).
Gauss’ law considers a hypothetical (imaginary) closed surface enclosing the charge distribution. This Gaussian
f h b h h h i i isurface can have any shape, but the shape that minimizes our calculations of the electric field is one that mimics () the symmetry of the charge distribution.the symmetry of the charge distribution.
If the charge is spread uniformly over a sphe e e enclose the sphe e ith aa sphere, we enclose the sphere with a spherical Gaussian surface and then find the electric field on the surface byy using the fact that
(:
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( )
2323--3 Flux of an Electric field3 Flux of an Electric field
To calculate how much charge is enclosed in Gauss’ law, , we need a way of calculating how much electricway of calculating how much electric field is intercepted by the Gaussian surface. This measure of interceptedp field is called flux ().
→
E||
A provisional definition for the flux of the electric field for the Gaussian surface is
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Then ∑ ∫→
2323--4 Gauss’ Law4 Gauss’ Law
Gauss’ law relates the net flux Φ of an electric field through a closed surface (a Gaussian surface) to the net charge qenc that is enclosed by that surface It tells us thatthat is enclosed by that surface. It tells us that
If qenc>0, Φ is outward; if qenc<0, Φ is inward.
Charge outside the Gaussian surface, no matter how large or how close it may be, is not included in the term qenc in Gauss’ law. The exact distribution of the charges
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qenc g inside the Gaussian surface are also of no concern.
The electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface,g , because as many field lines due to that charge enter the surface as leave it.surface as leave it.
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2323--5 Gauss’ Law and Coulomb’s Law 5 Gauss’ Law and Coulomb’s Law
From the spherical symmetry, at any point the electric field is also perpendicular to the surface andperpendicular to the surface and directed outward from the interior.
By Gauss’ law
2323--7 Applying Gauss’ Law: Cylindrical 7 Applying Gauss’ Law: Cylindrical SymmetrySymmetry
By symmetry, Gaussian surface ≡ a cylindrical surface + top d b tt i l f
SymmetrySymmetry
By Gauss’ law
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2323--8 8 Applying Gauss’ Law: Planar SymmetryApplying Gauss’ Law: Planar Symmetry
N d ti Sh tNonconducting Sheet
By symmetry, Gaussian surface ≡ a cylindrical surface + top and bottom circular surfaces
By Gauss’ law =•∫ encqAdE vv
0ε ∫
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2323--9 Applying Gauss’ Law: Spherical 9 Applying Gauss’ Law: Spherical SymmetrySymmetrySymmetrySymmetry
Here we can use Gauss’ law to prove the two shell theorems:
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(1) By Gauss’ law, we can obtain for r≥R
This field is the same as one set up by a point
This field is the same as one set up by a point charge q at the center of the shell of charge. Thus, the force produced by a shell of charge q on a charged particle placed outside theq on a charged particle placed outside the shell is the same as the force produced by a point charge q located at the center of the shell This proves the first shell theoremshell. This proves the first shell theorem.
(2) By Gauss’ law, we can obtain for r<R
Thus, if a charged particle were enclosed by the shell, the shell
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Thus, if a charged particle were enclosed by the shell, the shell would exert no net electrostatic force on the particle. This proves the second shell theorem.

30 3 43 Rπε
2323--9 A Charged Isolated Conductor9 A Charged Isolated Conductor
Some important theorems about conductors: (1) Charges can be found only on the surface of the conductor if the charges are in electrostatic equilibriumconductor if the charges are in electrostatic equilibrium.
This might seem reasonableThis might seem reasonable, considering that charges with the same sign repel one another. You
i h i i h b i hmight imagine that, by moving to the surface, the added charges are getting as far away from one another as they
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as far away from one another as they can.
(2) The electric field inside the conductor must be zero when the charges are in electrostatic equilibrium.e e c a ges a e e ec os a c equ b u
If this were not so, the field would exert forces on the conduction (free) electrons (always present in a conductor)( ) ( y p ) and thus current would always exist within a conductor. Of course, there is no such perpetual () current in an isolated conductor and so the internal electric field is zeroisolated conductor, and so the internal electric field is zero.
Ein=0
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(3) The electric field E at and just outside the conductor’s surface must also be perpendicular to that surface.su ace us a so be pe pe d cu a o a su ace
If it were not, then it would have a component along the conductor’s surface that would exert forces on the surface charges, causing them to move. However, such motion would violate our implicit assumption that we are dealing with electrostatic equilibrium Therefore E is perpendicularwith electrostatic equilibrium. Therefore, E is perpendicular to the conductor’s surface.
E
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00 εε ∫∫ ==• V
EA =
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Since the plates are conductors when we bring them into this
Two Conducting Plates
Since the plates are conductors, when we bring them into this arrangement, the excess charge on one plate attracts the excess charge on the other plate, and all the excess charge
t th i f f th l t With t i hmoves onto the inner faces of the plates. With twice as much charge now on each inner face, the new surface charge density (call it ) on each inner face is 2σ1.density (call it ) on each inner face is 2σ1.
σ∫∫ ==• Venc
dAq AdE vv
11 2EA2EA σ =→
+ =
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Ans: (a) qcentral=– 7.97×10−6 C ~ –8.0μC. (b) 11 5 10−6 C 12 C ( ) 5 3 C
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(b) qA=11.5×10−6 C ≈12 μC (c) –5.3 μC
Ans: x = 8.0 cm.
(b) x = − 0.691 m on the negative axis
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( ) g
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Ans: (a) E =0 (b) E =5.62×10-2 N/C (c) E = 0.112 N/C (d) E= 0.0499 N/C (d) E 0.0499 N/C (e) E= 0 (f) E= 0 ( ) Q 5 00 fC(g) Qi = –5.00 fC (h) Qo = 0
Ans: (a) E = 0 (b) E = 0 (c) E = 0 (d) E=7 32 N/C (c) E = 0 (d) E=7.32 N/C. (e) E=12.1 N/C. (f) E=1.35 N/C.
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Ans: (a) 2×10-9 C (b) -1.2×10-9 C
(c) In order to cancel the field (due to Q) within the conducting material, there must be an amount of charge equal to –Q distributed uniformly on the inner surface. Thus, the answer is +1.2×10-9 Cy ,
(d) Since the total excess charge on the conductor is q and is located on the surfaces, then the outer surface charge must equal the total minus the inner surface charge Thus the answer is 2 0×10−9 C
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minus the inner surface charge. Thus, the answer is 2.0×10 9 C – 1.2×10−9 C = +0.80 × 10−9 C.
Ans: (a) 0 (b) (c)
(d) Si E 0 f < th h th i f f th i(d) Since E = 0 for r < a the charge on the inner surface of the inner shell is always zero.The charge on the outer surface of the inner shell is therefore qa. Since E = 0 inside the metallic outer shell the net charge enclosed in a Gaussian surface that lies in between thenet charge enclosed in a Gaussian surface that lies in between the inner and outer surfaces of the outer shell is zero. Thus the inner surface of the outer shell must carry a charge –qa, leaving the charge on the outer surface of the outer shell to be qb+q
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charge on the outer surface of the outer shell to be qb+qa.
36 Ans: E=(5.65×104 N/C)
Ans: (a) A = –e/πa0 3 (b) ; points radially outward.
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