Electric Field of Uniformly Charged Solid Sphere
Radius of charged solid sphere: R Electric charge on sphere: 4 3 R . Q = V = 3 Use a concentric Gaussian sphere of radius r. Q r > R: E(4r2 ) =
0
E=
1 Q 4 0 r2 1 ,
4 3 r r < R: E(4r2 ) = 3 0
Faraday's Law of Induction (1)
Prototype: motional EMF reformulated. J Choose area vector A for current loop: A = Ls . R Magnetic flux: B = B dA. Here B = BLs. Motional EMF: E = vBL. Change in area of loop: dA = Lds. Change in magnetic flux: d
Particles Accelerated by Uniform Electric Field
A uniform electric field E = 0.75 103 N/C exists in the box. (a) A charged particle of mass m1 = 1.9 109 kg is released from rest at x = 3cm, y = 0. It exits the box at x = 3cm, y = 6cm after a time
Intermediate Exam I: Problem #1 (Spring '05)
The electric field E generated by the two point charges, 3nC and q 1 (unknown), has the direction shown. (a) Find the magnitude of E. (b) Find the value of q1 . Solution:
y E 45
o
3nC = 6.75N/C, (a) Ey =
Inductance of a Toroid
Total number of turns: N 0 I 2r Magnetic flux through each turn (loop): Z Z b 0 IN H b 0 IN H b dr = ln BH dr = B = 2 r 2 a a a Magnetic field inside toroid: B = 0 N 2 H b N B = ln Inductance: L I 2 a Narrow toroid: s b
Electric Potential of Conducting Spheres (1)
A conducting sphere of radius r1 = 2m is surrounded by a concentric conducting spherical shell of radii r2 = 4m and r3 = 6m. The graph shows the electric field E(r). (a) Find the charges q1 , q2 , q3 on th
Electric Potential of Charged Rod
y
Charge per unit length: = Q/L Charge on slice dx: dq = dx
x
dq = dx x
+ + + + + + + + + + +
dV
d
kdx kdq = x x
L
Electric potential generated by slice dx: dV = Electric potential generated by charged r
Electric Field of a Point Charge
electric charge
generates
electric field
exerts force locally
electric charge
exerts force over distance
(1) Electric field E generated by point charge q: E = k
q r ^ r2
(2) Force F1 exerted by field E on poi
Magnetic Field Application (16)
A currentcarrying wire consists of six straight segments as shown. J N Find the direction ( , ) and magnitude of the magnetic fields B1 , . . . , B6 produced by each segment. 0 I (sin 2  sin 1 ) and identify the quan
Atomic Structure of Matter
Periodic table: 100 elements. Building blocks of atoms: fundamental particles. particle electron proton neutron charge qe = e qp = +e qn = 0 mass me = 9.109 1031 kg mp = 1.673 1027 kg mn = 1.675 1027 kg
SI unit of
Kirchhoff's Rules
Loop Rule
When any closedcircuit loop is traversed, the algebraic sum of the changes in electric potential must be zero.
Junction Rule
At any junction in a circuit, the sum of the incoming currents must equal the sum of the outg
Electric Quadrupole Field
L +q p
2
x=0 2q
L +q p
1
E
E+
x
E
= =
kq kq k(2q) kq L L + + = 2 + 1+ 1 (x  L)2 (x + L)2 x2 x x x , , 2L 3L2 3L2 kq 2L 1+ + 2 + + 1  + 2  2 x2 x x x x 3kQ 6kqL2 = 4 x4 x (for x L)
q L +q L
",
2
,

Action and Reaction due to Coulomb Interaction
Two particles with masses m1 , m2 and charges q1 , q2 are released from rest a distance r apart. We consider the following four distinct configurations: (a) m1 = 1kg, m2 = 1kg, q1 = 1C, q2 = 1C (b) m1 =
Electric Potential Energy of Two Point Charges
Consider two different perspectives: q1 . #1a Electric potential when q1 is placed: V (r2 ) = V2 = k r12 q1 q2 Electric potential energy when q2 is placed into potential V2 : U = q2 V2 = k r12 q2 . #1b E
Electric Field from Electric Potential in Two Dimensions
Given is the electric potential: V (x, y) = ax2 + bxy 3 with a = 1V/m2 , b = 1V/m4 . Find the electric field: E(x, y) = Ex (x, y)^ + Ey (x, y)^ via partial derivatives. i j V = 2ax  by 3 ,
Ampre's law (Extended Version)
Conduction current: I. dE . Displacement current: ID = 0 dt I Ampre's law: B d = 0 (I + ID ) = 0 I + 0
dE . 0 dt
tsl315 p.1/1
Impedance Matching
A battery providing an emf E with internal resistance r is connected to an external resistor of resistance R as shown. For what value of R does the battery deliver the maximum power to the external resistor? Electric current: E 
Electric Potential of Charged Ring
Total charge on ring: Q Charge on arc: dq
Charge per unit length: = Q/2a
Find the electric potential at point P on the axis of the ring. dq kdq = r x 2 + a2 Z Z kQ k dq dq = V (x) = k = x 2 + a2 x 2 + a2
Particle Projected Perpendicular to Uniform Electric Field
A charged particle (m = 3kg, q = 1C) is launched at t0 = 0 with initial speed v0 = 2m/s in an electric field of magnitude E = 6 106 N/C as shown.
y m q 1m 1m
(a) Find the position of the pa
Ampre's Law (Restricted Version)
The circulation integral of the magnetic field B around any closed curve (loop) C is equal to the net electric current IC flowing through the loop: I with 0 = 4 107 Tm/A B d = 0 IC , H The symbol denotes an integr
Electric Dipole Potential
Use spherical coordinates: V = V (r, ) independent of azimuthal coordinate . , r  r+ (q) q = kq Superposition principle: V = V+ + V = k + r+ r r r+ Large distances (r L): r  r+ L cos , r r+ r2 V (r, ) k qL cos
Momentum Transport in Electromagnetic Plane Wave
The momentum transported by an electromagnetic wave is proportional to the energy transported. Momentum density: S p = 2, V c where S = 1 E B is the Poynting vector. 0
When the wave is absorbed by a
Intermediate Exam II: Problem #1 (Spring '05)
The circuit of capacitors connected to a battery is at equilibrium. (a) Find the equivalent capacitance Ceq . (b) Find the voltage V3 across capacitor C3 . (c) Find the the charge Q2 on capacitor C2 .
C3
Coulomb Force in One Dimension (1)
Find net force on charge q0 due to charges q1 and q2 . Consider xcomponent of force. F0 = +k q2 q0  q1 q0  k = +3.67 107 N  7.99 107 N = 4.32 107 N. 2 2 (3.5m) (1.5m)
Find net force on charge q2 due t
Velocity Selector
A charged particle is moving horizontally into a region with "crossed" uniform fields: an electric field E pointing down, a magnetic field B pointing into the plane. Forces experienced by particle: electric force F = qE pointing
Wave Equation
y(x, t) = A sin(kx  t) (displacement) v(x, t) = y = A cos(kx  t) (velocity) t 2y a(x, t) = 2 =  2 A sin(kx  t) (acceleration) t y = kA cos(kx  t) (slope of wave form) x 2y = k 2 A sin(kx  t) (curvature of wave form) x2 2