{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz2-solutions

# quiz2-solutions - 1 Quiz 2 solutions are in order of the...

This preview shows pages 1–2. Sign up to view the full content.

1 Quiz 2 solutions are in order of the questions for version 1. (1). Correct answer C For constant electric field, the potential difference is just the field times the distance, giving Δ V = (1500)(2 . 4) = 3600 V (2). Correct answer is D Here, the change in KE must equal negative the change in PE. Here, the distance traveled in 1.2m and hence the potential goes up by 1800V. Taking into account the definition of ev and the negative charge on the electron, this means that KE f = KE i q Δ V = 300(1 . 6 × 10 - 19 ) + (1 . 6 × 10 - 19 )(1800) = 3 . 4 × 10 - 16 (3). Correct answer is D This is a bit tricky. In the first part, the KE goes down from 1 2 mv 2 0 to 1 2 m ( v 0 2 ) 2 giving a change of KE of 3 8 mv 2 0 . By conservation of energy, this gives Ze 2 R = 3 8 mv 2 0 . When the proton has dropped to one quarter of the initial value, it has experienced a loss of KS which is now 15 32 mv 2 0 . (4). Correct answer is B In electrostatics, i.e. if there is no current, there is no potential difference between two objects connected by a wire.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}