1. PHY 2054: Physics II What is the Electric Field at the origin? What is the electric potential at the origin?

2. PHY 2054: Physics II What is the direction of Electric Field at the origin? What is the electric potential at the origin

3. PHY 2054: Physics II Calculate the Electric Field at P Calculate the el. potential at P

4. PHY 2054: Physics II Point charge Distribution of charges Line charge at an edge Disc on an axis through the center

5. PHY 2054: Class Quiz If 500 J of work are required to carry a charged particle between two points with a potential difference of 20V, the magnitude of the charge on the particle is: A. 0.040 C B. 12.5 C C. 20 C D. cannot be computed unless the path is given E. 25 C

6. Two particles with charges Q and -Q are fixed at the vertices of an equilateral triangle with sides of length a. The work required to move a particle with charge q from the other vertex to the center of the line joining the fixed particles is: A. 0 B. kQq/a C. kQq/a 2 D. 2kQq/a E. 1.4kQq/a q Q -Q

7. PHY 2054: Physics II a= 39 cm q 1 = 3.4pC q 2 = 6 pC E at the center? What is the potential at the center?

8. Capacitors and dielectrics Introduction of a dielectric  capacitance increases With battery connected  potential difference is fixed and maintained, charge increases. In other words, energy increases => dielectric is pushed out When the battery is not connected, charge stays the same, the energy decreases  dielectric is pulled in.

9. Problems

10. Here there are two capacitors in parallel

11. Videos How flash works http://electronics.howstuffworks.com/capacit or.htm

12. Here is one 16.37 2 4/3 The top serial combination has equivalent capacitance C1 = 2μF the bottom is C2 = 4/3 μF. The total equivalent Capacitance of the two in parallel is 2+4/3 = 10/3 μF. The charge on the outside plates on top is 2x90 = 180 μC =Q 6 = Q 3 On the bottom outside plates is 4/3x90 = 120 μC = Q 2 = Q 4 V 6 =potential difference across the 6 μF cap = Q/C = 30V V 3 = 60V V 2 = 60V and V 4 = 120/4 = 30V How much energy is stored on each capcitor? Q 2 /2C = CV 2 /2

13. 16.38 What is the equivalent capacitance? a. 38 Fb. 12 μFc. 20 μFd. 6 μF What is the total energy provided by the battery? How much energy is on the 4 μF capcitor? On the 2 μF capacitor? On the 24 μF capacitor? On the 8 μF capacitor? 7.78 mJ 2.59 mJ 1.30 mJ 0.97 mJ 2.92 mJ

14. Problems

15. 16.44

16. 16.61

17. ---------- = An infinite sequence of capacitors, each equal to C. Find the equivalent capacitance S. But watch… algebra-wise this is also described by the equation… Solve for S. = If you just look at the first link, by replacing s in the last arm by C, its equivalent capacitance is 1.33C, not a bad guess for the whole thing

18. If the areas are A 1 and A-A 1. C 123 = 2.4 μ F, q = 28.8 μ C C 2  C 24 = 12 μ F C 1234 = 3 μ F q =36 μ C C 1 = C 3 = 8 μF, C 2 = C 4 = 6 μF, V = 12V When the switch S is closed, how much charge passes through point P? How much passes through the switch S? Δq = 36-28.8 = 7.2 μC Q4 = 18μC

19. The last problem A 10 V battery is connected to a series of n capacitors. Each of the capacitance is 2 μF. If the total energy stored is 25 μJ, what is n? a. 4b. 25c. 10 d. 50e. None of these How much is the value of n, if the total energy stored is 10 μJ? a. 4b. 25c. 10 d. 50e. None of these