{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch23_p48 - is 8.0 × 10 7 N/C Therefore E shell E particle...

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

View Full Document Right Arrow Icon
48. Let E A designate the magnitude of the field at r = 2.4 cm. Thus E A = 2.0 × 10 7 N/C, and is totally due to the particle. Since E particle = q 4 πε o r 2 then the field due to the particle at any other point will relate to E A by a ratio of distances squared. Now, we note that at r = 3.0 cm the total contribution (from particle and shell)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: is 8.0 × 10 7 N/C. Therefore, E shell + E particle = E shell + (2.4 / 3 ) 2 E A = 8.0 × 10 7 N/C Using the value for E A noted above, we find E shell = 6.6 × 10 7 N/C. Thus, with r = 0.030 m, we find the charge Q using E shell = Q 4 πε o r 2 ¡ Q = 6.6 × 10 − 6 C ....
View Full Document

{[ snackBarMessage ]}