My office change was not reflected on the syllabus. It is now ESCN 2.206.

1. My office change was not reflected on the syllabus. It is now ESCN 2.206. Our first exam is a week from next Tuesday - Sep 27. It will cover everything I have covered in class including material covered next Thursday. There will be two review sessions Monday, Sep 26 - at 12:30 PM and at 3:00 PM in the same room as the problem solving session: FN 2.212. I have put several (37) review questions/problems on Mastering Physics. These are not for credit but for practice. I will review them at the review session Monday.

2. Potential of a point charge Moving along the E-field lines means moving in the direction of decreasing V. As a charge is moved by the field, it loses potential energy, whereas if the charge is moved by the external forces against the E-field, it acquires potential energy

3. Negative charges are a potential minimum Positive charges are a potential maximum

4. Positive Electric Charge Facts • For a positive source charge • Electric field points away from a positive source charge • Electric potential is a maximum • A positive object charge gains potential energy as it moves toward the source • A negative object charge loses potential energy as it moves toward the source

5. Negative Electric Charge Facts • For a negative source charge • Electric field points toward a negative source charge • Electric potential is a minimum • A positive object charge loses potential energy as it moves toward the source • A negative object charge gains potential energy as it moves toward the source

6. Electron Volts Electron volts – units of energy 1 eV – energy a positron (charge +e) receives when it goes through the potential difference Vab =1 V Unit: 1 Volt= 1 Joule/Coulomb (V=J/C) Field: N/C=V/m 1 eV= 1.6 x 10-19 J Just as the electric field is the electric force per unit charge, the electrostatic potential is the potential energy per unit charge.

7. Clicker question There is a 12 V potential difference between the positive and negative ends of a set of jumper cables, which are a short distance apart. An electron at the negative end ready to jump to the positive end has a certain amount of potential energy. On what quantities does this electrical potential energy depend? a. the distance and the potential difference between the ends of the cables b. the distance and the charge on the electron c. the potential difference and the charge d. the potential difference, charge, and distance

8. Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a negative ion with a charge at this terminal. What are the electric potential and electric potential energy of the negative ion relative to the electron? a. The electric potential and the electric potential energy are both twice as much. b. The electric potential is twice as much and the electric potential energy is the same. c. The electric potential is the same and the electric potential energy is twice as much. d. The electric potential and the electric potential energy are both the same. e. The electric potential is the same and the electric potential energy is increased by the mass ratio of the oxygen ion to the electron.

9. Examples A small particle has a charge -5.0 mC and mass 2*10-4 kg. It moves from point A, where the electric potential is fa =200 V and its speed is V0=5 m/s, to point B, where electric potential is fb =800 V. What is the speed at point B? Is it moving faster or slower at B than at A? E A B F In Bohr’s model of a hydrogen atom, an electron is considered moving around a stationary proton in a circle of radius r. Find electron’s speed; obtain expression for electron’s energy; find total energy.

10. Calculating Potential from E field • To calculate potential function from E field

11. When calculating potential due to charge distribution, we calculate potential explicitly if the exact distribution is known. If we know the electric field as a function of position, we integrate the field. Generally, in electrostatics it is easier to calculate a potential (scalar) and then find electric field (vector). In certain situation, Gauss’s law and symmetry consideration allow for direct field calculations. Moreover, if applicable, use energy approach rather than calculating forces directly (dynamic approach) Example: Solid conducting sphere Outside: Potential of the point charge Inside: E=0, V=const

12. Potential of Charged Isolated Conductor • The excess charge on an isolated conductor will distribute itself so all points of the conductor are the same potential (inside and surface). • The surface charge density (and E) is high where the radius of curvature is small and the surface is convex • At sharp points or edges  (and thus external E) may reach high values. • The potential in a cavity in a conductor is the same as the potential throughout the conductor and its surface

13. At the sharp tip (r tends to zero), large electric field is present even for small charges. Corona – glow of air due to gas discharge near the sharp tip. Voltage breakdown of the air Lightning rod – has blunt end to allow larger charge built-up – higher probability of a lightning strike

14. Example: Potential between oppositely charged parallel plates From our previous examples Easy way to calculate surface charge density Remember! Zero potential doesn’t mean the conducting object has no charge! We can assign zero potential to any place, only difference in potential makes physical sense

15. Vb = 0 – not a good choice as it follows Why so? Example: Charged wire We already know E-field around the wire only has a radial component We would want to set Vb = 0 at some distance r0 from the wire r - some distance from the wire

16. Example: Sphere, uniformly charged inside through volume Q - total charge - volume density of charge This is given that at infinity j =0

17. Potential Gradient We can calculate potential difference directly Components of E in terms of j Frequently, potentials (scalars!) are easier to calculate: So people would calculate potential and then the field Superposition for potentials: V = V1 + V2 + …

18. Example: A positively charged (+q) metal sphere of radius ra is inside of another metal sphere (-q) of radius rb. Find potential at different points inside and outside of the sphere. -q 1 a) 2 +q Total V=V1+V2 b) c) Electric field between spheres