{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CH 16 - Electric Potential-solutions

# CH 16 - Electric Potential-solutions - This print-out...

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

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.

Unformatted text preview: This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 6 % of the speed of light (2 . 998 × 10 8 m / s), starting from rest? The mass of an electron is 9 . 109 × 10- 31 kg and the charge on an electron is 1 . 602 × 10- 19 C. Ignore special relativity for this problem. Correct answer: 919 . 907 V. Explanation: Let : s = 6% = 0 . 06 , c = 2 . 998 × 10 8 m / s , m e = 9 . 109 × 10- 31 kg , and q e = 1 . 602 × 10- 19 C . The speed of the electron is v = 0 . 06 c = 0 . 06 ( 2 . 998 × 10 8 m / s ) = 1 . 7988 × 10 7 m / s , By conservation of energy 1 2 m e v 2 =- (- q e ) Δ V Δ V = m e v 2 2 q e = ( 9 . 109 × 10- 31 kg ) × ( 1 . 7988 × 10 7 m / s ) 2 2 (1 . 602 × 10- 19 C) = 919 . 907 V . 002 10.0 points Particles A (of mass m and charge Q ) and B (of m and charge 5 Q ) are released from rest with the distance between them equal to 1 . 8969 m. If Q = 29 μ C, what is the kinetic energy of particle B at the instant when the particles are 3 . 8969 m apart?...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

CH 16 - Electric Potential-solutions - This print-out...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online