Force and Torque on Electric Dipole
The net force on an electric dipole in a uniform electric field vanishes. However, this dipole experiences a torque = p L that tends to align the vector p with the vector E. Now consider an electric dipole tha
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
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
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 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
,

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
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: Magnetic Field Inside a Long Solenoid
Apply Ampre's law, I B d = 0 IC , to the rectangular Amperian loop shown.
Magnetic field inside: strong, uniform, directed along axis. Magnetic field outside: negligibly weak. Number of turns pe
Electric Field of Charged Semicircle
Consider a uniformly charged thin rod bent into a semicircle of radius R. Find the electric field generated at the origin of the coordinate system. Charge per unit length: = Q/R Charge on slice: dq = Rd (assume
Current Produced by Motional EMF
Motional EMF: E = vBL Terminal voltage: Vab = E  Ir E Electric current: E  Ir  IR = 0 I = r+R Applied mechanical force: Fapp Magnetic force: FB = I L B Motion at constant velocity: Fapp = FB Electrical po
Ampre's Law: Coaxial Cable
Consider a long coaxial cable, consisting of two cylindrical conductors separated by an insulator as shown in a crosssectional view. There is a current I flowing out of the plane in the inner conductor and a current of equ
Battery with Internal Resistance
Real batteries have an internal resistance r. The terminal voltage Vba Va  Vb is smaller than the emf E written on the label if a current flows through the battery. Usage of the battery increases its internal res
Unit Exam II: Problem #1 (Spring '07)
Consider the configuration of two point charges as shown. (a) Find the energy U3 stored on capacitor C3 . (b) Find the voltage V4 across capacitor C4 . (c) Find the voltage V2 across capacitor C2 . (d) Find the c
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
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
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
Hall Effect
Method for dermining whether charge carriers are positively or negatively charged. Magnetic field B pulls charge carriers to one side of conducting strip. Accumulation of charge carriers on that side and depletion on opposite side produ
Intermediate Exam III: Problem #1 (Spring 05)
An infinitely long straight current of magnitude I = 6A is directed into the plane () and located a
distance d = 0.4m from the coordinate origin (somewhere on the dashed circle). The magnetic
~ generated by th
Unit Exam IV: Problem #1 (Spring 12)
In the circuit shown we close the switch S at time t = 0. Find the current IL through the inductor
and the voltage V6 across the 6resistor
(a) immediately after the switch has been closed,
(b) a very long time later.
Intermediate Exam III: Problem #1 (Spring 05)
An infinitely long straight current of magnitude I = 6A is directed into the plane () and
located a distance d = 0.4m from the coordinate origin (somewhere on the dashed
~ generated by this current is in the n
PHY204 UNIT EXAM 2, section 2 (Spring 2016)
Problem 1
The capacitor circuit shown is connected to a 12 V battery. Find
(a) the equivalent capacitance Ceq ;
(b) the electric charge Q4 on the 4 F capacitor;
(c) the potential energy U3 stored on the 3 F capa
PHY204 UNIT EXAM 3, section 2 (Spring 2016)
Problem 1
(a) A point charge q = 3.5 nC moves along the positive zaxis through
a uniform magnetic field B = 0.55 i T at a velocity v = 2.0 107 m/s. Find
the force Fa (all components of this vector) acting on th