P23_057 - 57. We note that the contributions to the field...

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

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

Unformatted text preview: 57. We note that the contributions to the field from the pair of −2q charges exactly cancel, and we are left with the (opposing) contributions from the 4q (at r = 2d) and −q (at r = d) charges. Therefore, using k = 1/4πε0 4q q |Enet | = k −k 2 =0 . 2 (2d) d The net field at P vanishes completely. ...
View Full Document

Ask a homework question - tutors are online