1. Work done by electric force (source: fixed charges) on a test charge depends only on endpoints, not on path. (You can see this easily for a single fixed charge… it holds in general because of superposition.) Electric forces are “conservative” - We can define a potential energy. When a + charge moves “down the field”, the electric force does work on it, increasing its kinetic energy (or putting energy elsewhere). When a + charge moves “up the field”, it either loses kinetic energy, or some other force must push it up.

2. Example : a charge compressing a spring.

3. The electrical potential energy of a system of charges is the work necessary to assemble the charges from “infinity”. (For point charges, we take U=0 at infinity.) This will include all pairs of interactions. Two equal + charges are initially stationary and separated by r 0. If they are allowed to fly apart (to infinity), what will be the kinetic energy of each? A]B] C]D]

4. Three equal + charges are initially stationary and at the vertices of an equilateral triangle with side r 0. If they are allowed to fly apart (to infinity), what will be the kinetic energy of each? A]B] C]D]

5. Just as we can define electric field as the force felt by a test charge We define “potential” as potential energy of a test charge. Just as a conservative force is: (minus) the derivative of the potential energy The electric field is (minus) the derivative of the potential.

7. Note: just because V=0 doesn’t mean E=0! A function can be zero but have a non-zero derivative. Also: it’s time to think in 3D. The derivative can be taken w.r.t x, y, or z. This means: hold y, z constant, so dy=dz=0 Note that E is a vector, but V is a scalar.

8. Rank the magnitudes (smallest to largest) of the electric field at point P in the three arrangements shown. ANSWER=D A] all are the same B] I, II, III C] III, II, I D] II, I, III Rank the electric potentials at point P (smallest to largest). ANSWER=A

9. If V = -4x 2 + 4y (x,y in meters, V in volts), what is the E field at the origin? ANSWER=C A] 0 B] 4 V/m in the +y direction C] 4 V/m in the -y direction D] 4 V/m directed at an angle between +x and -y axes. E] cannot determine

10. If V = -4x 2 + 4y (x,y in meters, V in volts), where is the E field = 0? ANSWER=E A] at the origin B] at x=1, y=1 C] at x=-1, y=1 D] at both B and C E] nowhere.

11. Equipotential Lines = Contours of constant V E field points downhill Downhill is always perpendicular to level Conductors at rest are equipotential