1. (10) Find the conventional current through each resistor in the circuit shown below. Be sure to indicate the direction in which each current flows.
   
   

2. (10) A wire has been bent into the shape shown in the diagram below. The curved portion is three-quarters of a circle. The radius of the curved piece is R and the wire carries a linear charge density λ. The straight pieces each have length 5R. Find the electrostatic potential at point P at the centre of the circular piece. Hint - consider each piece separately.
   
   

3. (10) In the circuit below, V = 20.0 Volts, C₁ = 6.00 μF, C₂ = 1.00 μF, C₃ = 4.00 μF, and C₄ = 4.00 μF.
   (a) Switch S₁ is closed and C₁ is allowed to charge fully. What is the charge on C₁?
   (b) Switch S₁ is now reopened and then switch S₂ is closed. What is the final charge on each capacitor? (Hint - first consider the equivalent capacitor C₂₃₄)
   
   

4. (10) In the diagram below are two wires both with charge density λ. One piece is bent into a semicircle of radius R. The other is an infinitely long ( L >> R ) wire.
   (a) What is the electric field at point P due solely to the semicircular wire? Point P is the centre of the full circle.
   (b) What is the electric field at P due to the long straight wire? (You may use Gauss' Law to derive this result if you wish.)
   (b) What should a be for there to be no net field at point P?
   
   

5. (15) An insulating sphere of radius R has a volume charge density given by
   
   ρ = ρ₀(a - r/b)

   where ρ₀, a, and b are positive constants and r is the radial distance from the centre of the sphere.
   (a) Determine the magnitude of the electric field for (i) r < R and (ii) r > R. Sketch the results.
   (b) Use the result from part (a) the determine the potential difference between r = 0 and r = R. Which point is at the higher potential?
   

6. (10) In the circuit below, V = 20.0 Volts, C₁ = 6.00 μF, and C₂ = 3.00 μF. Consider the following possibilities. Each case starts with the switches opened and the capacitors discharged.
   (a) Switch S₁ is closed and C₁ is allowed to charge fully. Next switch S₁ is now reopened and then switch S₂ is closed. What is the final charge on each capacitor?
   (b) A dielectric of k = 5.00 is put between the plates of a discharged C₁. Switch S₁ is closed and C₁ is allowed to charge fully. Next switch S₁ is now reopened and then switch S₂ is closed. What is the final charge on each capacitor?
   (c) Switch S₁ is closed and C₁ is allowed to charge fully. Next switch S₁ is now reopened and then a dielectric of k = 5.00 is placed between the plates of a discharged C₁. Then switch S₂ is closed. What is the final charge on each capacitor?
   
   

Coulomb's Law and Electric Fields:

F = kq1q2/r2	F= q0E	EPoint Charge = kQ/r2
Ewire = 2kλ/r	Eplate = Σ/2ε0	Esphere = KQ/r2

Gauss' Law

	φE = òE ·ndA	= Qinside/ε0
cylinders -	E(a)2πaL	= [2πL òr(r)rdr]/ε0
spheres -	E(a)4πa2	= [4pòr(r)r2dr]/ε0

Electrostatic Potential

ΔV = V(b) - V(a) = -òbaE·dl = -òbaE(r)dr
Vpoint charge = kQ/r	V = V1 + V2 + ...	ΔK =ΔU = -qΔV

Capacitors

C = Q/V	CP = C1+C2	1/Cs = 1/C1+1/C2
Cplates = ε0A/d	Eplates = Q/ε0A	Cdielectric = kC0
Edielectric = kE0	Σb = [(k-1)/k]sf	U = ½Q2/C = ½QV = ½CV2
charging:	Q = Cε(1 - e-t/RC)	I = (ε /R)e-t/RC
discharging:	Q = Q0e-t/RC	I = (ε /R)e-t/RC

Resistors

Rs = R1 + R2	1/Rp = 1/R1 + 1/R2
V = IR	P = I2R = V2/R = VI
åEi -åIiRi = 0	åqin =åqout

Constants

k = 8.99 × 109 N m2/C2

ε0 = 8.85 × 1012 F/m

μ0 = 4π × 107 T-m/A

Identities and Integrals

if ax2+bx+c=0, x = {b±[b2- 4ac]½}/2a	S = Rθ, dS = Rdθ
òxndx = xn+1/(n+1) + C	òx-1dx = lnx + C
òebxdx = ebx/b + C	òxebxdx = (x - b)ebx + C
òcosΩxdx = (sinΩx)/Ω + C	òsinΩxdx = -(cosΩx)/Ω + C
ò(a/[a2+x2]3/2)dx = 1/[a[a2+x2]1/2] + C	ò(x/[a2+x2]3/2)dx = -1/[a2+x2]1/2 + C

Questions? mike.coombes@kpu.ca