A different review A review of the following topics by working on problems 1.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 dipole Calculate the electric field of a point on the x and y axis, far away from an electric dipole. x x P Formula to use: For a point on x:

Calculate the electric field of a point on the x and y axis, far away from an electric dipole. x x P For a point on y, the field y components cancel.

Electric field of a charged rod Calculate 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. O Formula to use: Because 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. Choose a small section on the rod: The electric field at point O:

Force/field equilibrium Two 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? Formula to use: Condition for equation: equilibrium on the third bead. Special case: 1-D problem. Construct the x-coordinate system. O x Construct equation: q3q3 solution:discard: (Why?)

stable equilibrium ? Translate: prove that the electric potential energy of the whole system when bead 3 at position x is the minimum. O x q3q3 Change bead 3 by, the change in potential energy is Formula to use: Answer: bead 3 at stable equilibrium.

Gauss’s Law a type X problem? A sphere of radius 2 a 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: See example in lecture note

Now the electric field of the material of the cavity:

Some easier problems Finding the C eq But So Solve the is equation, we get

Some easier problems Finding the C eq Wire has no resistance. Re-draw the diagram: 4.0 μF 6.0 μF 5.0 μF 7.0 μF a b So

Kirchhoff’s rules Determine the current in each branch. L1 L2 I1I1 I3I3 I2I2 Step 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: Step 5: simplify and solve the three equations for the three variables. + – – – – – +