ch26-p070 - towards the negative terminal (low potential),...

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

View Full Document Right Arrow Icon
70. (a) The current is 4.2 × 10 18 e divided by 1 second. Using e = 1.60 × 10 19 C we obtain 0.67 A for the current. (b) Since the electric field points away from the positive terminal (high potential) and
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: towards the negative terminal (low potential), then the current density vector (by Eq. 26-11) must also point towards the negative terminal....
View Full Document

Ask a homework question - tutors are online