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Unformatted text preview: Chapter 23: Gauss’ Law PHY2049: Chapter 23 1 Electric Flux
Simple definition of electric flux (E constant, flat surface)
E at an angle to planar surface, area A Normal Units = N m2 / C (SI units) E Simple example
Let E = 104 N/C pass through 2m 5m rectangle, 30° to normal 4 E = 10 * 10 * cos(30) = 100,000 * 0.866 = 86,600 More general E definition (E variable, curved surface) PHY2049: Chapter 23 2 Example of Constant Field PHY2049: Chapter 23 3 Flux Through Closed Surface
Surface elements dA always point outward! Sign of
E outward (+) E inward ( )
E E <0 E >0 E =0 PHY2049: Chapter 23 4 Example: Flux Through Cube
E field is constant:
Flux Flux Flux Flux through through through through front face? + EA back face? EA 0 top face? whole cube? 0 PHY2049: Chapter 23 5 Example: Flux Through Cylinder
Assume E is constant, to the right Flux through left face? EA Flux through right face? + EA
Flux through curved side 0 Total flux through cylinder? 0 PHY2049: Chapter 23 6 Example: Flux Through Sphere
Assume point charge +Q E points radially outward (normal to surface!) Foreshadowing of Gauss’ Law!
PHY2049: Chapter 23 7 Gauss’ Law
General statement of Gauss’ law
Integration over closed surface qenc is charge inside the surface Charges outside surface have no effect Can be used to calculate E fields. But remember
Outward E field, flux > 0 Inward E field, flux < 0 Consequences of Gauss’ law (as we shall see)
E = 0 inside conductor E is always normal to surface on conductor Excess charge on conductor is always on surface Conductor PHY2049: Chapter 23 8 Reading Quiz
What is the electric flux through a sphere of radius R surrounding a charge +Q at the center?
1) 2) 3) 4) 5) Q +Q/ 0 Q/ 0 +Q None of these PHY2049: Chapter 23 9 Question
dS
+Q +Q dS 1 How does the flux E through the entire surface change when the charge +Q is moved from position 1 to position 2? a) b) c)
E E E 2 increases decreases doesn’t change
PHY2049: Chapter 23 Just depends on enclosed charge, not its location
10 Example of Flux
Charges on shells are
+Q (ball at center) +3Q (middle shell) +5Q (outside shell) Find fluxes on the three Gaussian surfaces (a) Inner +Q/ 0 (b) Middle +4Q/ 0 (c) Outer +9Q/ 0 PHY2049: Chapter 23 11 Power of Gauss’ Law: Calculating E Fields
Valuable for cases with high symmetry
E = constant, surface E  surface Spherical symmetry examples
E field vs r for point charge E field vs r inside uniformly charged sphere Charges on concentric spherical conducting shells Cylindrical symmetry examples
E field vs r for line charge E field vs r inside uniformly charged cylinder Rectangular symmetry examples
E field for charged plane E field between conductors, e.g. capacitors
PHY2049: Chapter 23 12 Example
4 Gaussian surfaces: 2 cubes and 2 spheres
Rank magnitudes of E field on surfaces Which ones have variable E fields? What is the flux E over each the Gaussian surface (a) E falls as radius increases (b) E is nonconstant on cube (r changes) (c) E is same for each case, E = +Q/ 0 PHY2049: Chapter 23 13 Derive Coulomb’s Law From Gauss’ Law
Charge +Q at a point
Spherical symmetry: E must be radially symmetric r Draw Gaussian’ surface around point
Sphere of radius r E is constant, to Gaussian surface S E dA = E 4 r ()
2 = Q
0 Gauss’ Law Gaussian surface (sphere) Solve for E PHY2049: Chapter 23 14 Another Spherical Symmetry Example
Charged sphere with constant charged density
Find E vs distance r Two cases: (1) r < R and (2) r > R Spherical symmetry: E is radial 4 Q= R3 3 Case (1): Again pick spherical Gaussian surface E dA = qenc
0 Gauss’ law
2 R r S S E dA = E 4 r
4 3 () E constant, radial Enclosed charge qenc = r3 PHY2049: Chapter 23 15 Spherical Symmetry (cont)
So Gauss’ law gives us E 4 r 2 = Rewrite, using total charge Q = E= Q 4
0R 2 4 3 () 4 3 r3
0 E= 3 r
0 R3 r R r E0 R for r < R Case 2: r > R, qenc = Q E= Q 4
0r 2 E 4 r2 =
for r > R () Q
0 Solutions are equal at r = R!
PHY2049: Chapter 23 E0 = Q 4 R2 0
16 Plot of E(r) vs r
E= E Q 4 R2 0 r R E= Q 4 r2 0 O R
PHY2049: Chapter 23 r
17 General Use of Spherical Symmetry
For any charge distribution depending only on radius:
No preferred direction spherical symmetry E is radial and depends only on radius r (r) LHS of Gauss’ law: S E d A = E 4 r2 dV =
r 0 () RHS of Gauss’ law: qenc = qenc 4 r2 V r (r ) 4 r 2 dr E= = 0 (r ) 4
4 r2 r 2 dr So we can calculate E(r) for complicated
PHY2049: Chapter 23 (r)!
18 Another Example
Let charge density (r) be ( r ) = Kr
r 0 r R Find E inside the sphere: r < R
S E dA = E 4 r K +3
+1 ()
2 qenc = Kr E= ( 4K 4 r dr = r +3
2 ) +3 E=
For ( ) r
0 Rewrite as Q 4 R2 0 r R +1 =0, we recover the case for (r) = const. Outside the sphere: qenc = Q E= Q 4 r2 0
PHY2049: Chapter 23 19 Cylindrical Symmetry
Consider a charged rod with uniform
Infinite length assumed By axial symmetry, E field must be radial = charge/length Find E(r)
Pick Gaussian surface to be cylinder, radius r
S S E dA = qenc
0 Gauss’ law E d A = E 2 rh ( ) LHS RHS qenc = h E= 2 1 r 0 PHY2049: Chapter 23 20 Compare to Solution for Finite Rod
From direct integration (Ch. 22), E field at distance r above center of wire of length L is E= 1 2
0r 1 + 4r 2 / L2 This becomes our solution on previous page when L =
Let L = 10, r = 0.5: Gauss’ law too large by ~0.5% Let L = 10, r = 1.0: Gauss’ law too large by ~2% Let L = 10, r = 2.0: Gauss’ law too large by ~8% So Gauss’ law solution is fairly accurate as long as r is small compared with L and not close to the endpoints.
PHY2049: Chapter 23 21 Example
Given a cylinder of height 2 m and radius 20 cm.
How much charge can it hold before field at surface causes a breakdown? Breakdown in air occurs when E = 3 10 6 N/C = q/L q/ L 1 E= 2 0r q=2 0 LrE = 6.28 ( ) ( 8.85 10 12 )(2)(0.2)(3 106 )
22 = 67 μC PHY2049: Chapter 23 ...
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