A different reviewA review of the following topics by working on problems1. Electric charge, potential, force, field and flux. Coulomb's Law Gauss’s Law.2. Capacitance and resistance. Ohm’s Law.3. Circuits. Kirchhoff’s Rules.Electric field of a dipoleCalculate the electric field of a point on the x and y axis, far away from an electric dipole. xxPFormula to use:2ˆeeoqkq r= =FE rFor a point on x:( )a+=+2eqE kx( )a−−=−2eqE kx( ) ( )( ) ( )( ) ( )( ) ( )( )a aa aa aa a-2x2aaa-4a+ −−= + = ++ − = − + − − − += − = + −+ −=−=∵∵ 2 22 22 22 22 222 231 1 ( )( ) x e ex eeeeq qE E E k kx xE k qx xx xk q x y x y x yx xk q xxqkxCalculate the electric field of a point on the x and y axis, far away from an electric dipole. xxPFor a point on y, the field y components cancel. ( )aaaaaaa= =++=+= ∵ 322 22 22 232 cos cos2 2 x eeeqE k θ θyyqkyqk yyElectric field of a charged rodCalculate the electric field at the origin O of the coordinator system, generated by an evenly charged rod with total charge +Q, with length l, and placed distance a away from O on the x-axis. OFormula to use:2ˆeeoqkq r= =FE rBecause all the charge is concentrated on the x-axis, the electric field has only the x component, pointing to negative x direction at point O.= −2x eλdxdE kxChoose a small section on the rod:The electric field at point O:( )a laa la laaa a + l+++= = − = − − = − ∫∫21x xex e eE dEk QdxE k λ k λx xForce/field equilibriumTwo small positively charged beads are fixed at the ends of a horizontal insulating rod. A third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Is this a stable equilibrium?1 22eq qkr=12 12F rFormula to use:Condition for equation: equilibrium on the third bead.Special case: 1-D problem.Construct the x-coordinate system.OxConstruct equation:( )3 3223e eq q q qk kxd x⋅ ⋅=−q3solution:33 1x d=+33 1x d=−discard:(Why?)stable equilibrium?Translate: prove that the electric potential energy of the whole system when bead 3 at position x is the minimum. Oxq3Change bead 3 byx∆, the change in potential energy is()( )( )( )( )( )( )3 333333 233 2 this term 0so follow the sign of When 0 >0, 0 means is minimumWhen 0 0, 0 means e e ed x xq q q qU k k k q qx x d x x x x d x xd x xd x,x x d x xU q .q , U U U xq , U U U∆∆∆ ∆ ∆ ∆∆∆ ∆∆∆∆ − +⋅ ⋅= + = ⋅ + − − + − − − +> > + − − > >< < <∵( ) is minimumxFormula to use:1 2eq qU kr⋅=Answer: bead 3 at stable equilibrium.Gauss’s Lawa type X problem?A sphere of radius 2a is made of insulating material that has uniform volume charge density ρ. Assume that the material itself does not affect the electric field (i.e., ignore its dielectric effect). A spherical cavity of radius a is made in the original sphere as shown in the figure. Calculate the electric field in the cavity.It is a Gauss’s Law problem. Put the material back to the cavity, the problem is type I. The electric field inside the sphere is: 03ρε=E rSee example in lecture noteNow the electric field of the material of the cavity:( )003where is in the system its center is the center of the cavity.Moving back to the original x-y-z coordinate system: so 3the answer, electric field in the cavitcm ccc cmˆ ˆaj - ajρερε== + = E rrr r E r( )0 0 00y:3 3 3or: 0 03 c cmcx cy czˆ ˆ- - aj ajE , E a, Eρ ρ ρε ε ερε= − = == = = E E E r rcrrSome easier problemsFinding the Ceq()( )10 0210 02ABXYABC C CCC C C+=+ +ButSoXY ABC C C= ≡()( )10 0210 02C C CCC C C+=+ +Solve the is equation, we get2 20 0002 2 03 1the answer is 2discard the other solution 3 1 2C C C CC CC C+ − =−=+= −Some easier problemsFinding the Ceq5 0 7 04 0 6 0 F 12.9 F5 0 7 0ab. .C . .. .µ µ×= + + ≅+Wire has no resistance. Re-draw the diagram:4.0 µF6.0 µF5.0 µF7.0 µFabSoKirchhoff’s rulesDetermine the current in each branch.L1L2I1I3I2Step 1: choose and mark the loop.Step 2: choose and mark current directions. Mark the potential change on resistors.Step 3: apply junction rule:Step 4: apply loop rule:− − + =1 2 30I I I. . .. . . . .+ − =+ − + − =3 13 2L1: +4.00-(100 5 00) 8 00 0L2: +4.00-(100 5 00) (3 00 100) 12 0 0I II IStep 5: simplify and solve the three equations for the three variables.+––––+++–++ − =+ − =+ + =1 2 31 32 304 3 2 02 3 4 0I I II II I...== −= −1231 50 A8 50 A7 00
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