You have a lot of POTENTIAL...but you need to WORK harder.

p. 522, Q7, P 4,5,6,8,9

Question 7. Electric potential is the amount of energy per unit charge, while electric field is the amount of electric FORCE per unit charge. Electric potential is just electric potential energy divided by charge.

Problems.
4. If the electron gains 3.6 x 10^-16 J of KE, it means it loses a potential energy of 3.6 x 10^-16 J. This happens as it crosses a potential difference delta_V = delta_U/q.

Thus delta_V = (3.6 x 10^-16 J)/(-1.6 x 10^-19 C) = 2,156 V.

Since it is an electron with a negative charge, the final plate (plate B) is at the higher potential.

5. delta_V = Ed, E = delta_V/d = 220 V/(.0052 m) = 4.23 x 10⁴ N/C or V/m.

6. Following the same procedure as with #5, the answer is 7.04 V.

8. d = delta_V / E = 3.3 x 10^-5 m

9. Since we have a non-conservative force doing work, a change in kinetic energy, and potential difference, it sounds like we might be able to use the generalized work energy theorem.

W_ncf = delta_K + delta_U
25 x 10^-4 J = 1.82 x 10^-4 J + delta_U
delta_U = 2.02 x 10^-3 J
delta_V = delta_U/q = 2.02 x 10^-3 J/(-7.5 x 10^-6 C = -269 V. This is the potential difference from a to b for the negative charge. The potential difference Va - Vb (the reverse of Vb - Va) = +269 V.