5 7 pm in ACS 130 Midterm Friday, February 14 No tutorials this - - PowerPoint PPT Presentation

❑ QSCU Midterm Review session – Wednesday, February 12 (today) ➢ 5 – 7 pm in ACS 130 ❑ Midterm – Friday, February 14  ❑ No tutorials this week ❑ No homework this week

This is a graph of the x-component

• f the electric field along

the x-axis. The potential is zero at the origin. What is the potential at x = 1m?

QuickCheck 26.1

A. 2000 V B. 1000 V C. 0 V D. E. –1000 V –2000 V

This is a graph of the x-component

• f the electric field along

the x-axis. The potential is zero at the origin. What is the potential at x = 1m?

QuickCheck 26.1

ΔV = –area under curve

A. 2000 V B. 1000 V C. 0 V D. E. –1000 V –2000 V

At which point is the electric field stronger?

QuickCheck 26.2

• A. At xA
• B. At xB
• C. The field is the same strength at both.
• D. There’s not enough information to tell.

At which point is the electric field stronger?

QuickCheck 26.2

• A. At xA
• B. At xB
• C. The field is the same strength at both.
• D. There’s not enough information to tell.

|E| = slope of potential graph

An electron is released from rest at x = 2 m in the potential

• shown. What does the electron do right after being released?

QuickCheck 26.3

• A. Stay at x = 2 m
• B. Move to the right (+ x) at steady speed.
• C. Move to the right with increasing speed.
• D. Move to the left (– x) at steady speed.
• E. Move to the left with increasing speed.

QuickCheck 26.3

Slope of V negative => Ex is positive (field to the right). Electron is negative => force to the left. Force to the left => acceleration to the left.

An electron is released from rest at x = 2 m in the potential

• shown. What does the electron do right after being released?
• A. Stay at x = 2 m
• B. Move to the right (+ x) at steady speed.
• C. Move to the right with increasing speed.
• D. Move to the left (– x) at steady speed.
• E. Move to the left with increasing speed.