# ch24_p62 - evaluated at P Once we know the electric field...

This preview shows page 1. Sign up to view the full content.

6 2. (a) When the proton is released, its energy is K + U = 4.0 eV + 3.0 eV (the latter value is inferred from the graph). This implies that if we draw a horizontal line at the 7.0 Volt “height” in the graph and find where it intersects the voltage plot, then we can determine the turning point. Interpolating in the region between 1.0 cm and 3.0 cm, we find the turning point is at roughly x = 1.7 cm. (b) There is no turning point towards the right, so the speed there is nonzero, and is given by energy conservation: v = 2(7.0 eV) m = 2(7.0 eV)(1.6 x 10 -19 J/eV) 1.67 x 10 -27 kg = 20 km/s. (c) The electric field at any point P is the (negative of the) slope of the voltage graph
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: evaluated at P . Once we know the electric field, the force on the proton follows immediately from F → = q E → , where q = + e for the proton. In the region just to the left of x = 3.0 cm, the field is E → = (+300 V/m) ˆ i and the force is F = +4.8 × 10 − 17 N. (d) The force F G points in the + x direction, as the electric field E G . (e) In the region just to the right of x = 5.0 cm, the field is E → =(–200 V/m) ˆ i and the magnitude of the force is F = 3.2 × 10 − 17 N. (f) The force F G points in the − x direction, as the electric field E G ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online