workout 24 - HPHY 253 Workout 24 Gauss&s Law for...

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

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: HPHY 253 Workout 24 Gauss&s Law for Electricity Dr. A. E. Bak Name 1. Serway & Jewett Problem 24 : 4 . Consider a closed triangular box resting within a horizontal electric &eld of magnitude E = 7 : 80 & 10 4 N= C as shown in Figure P 24 : 4 . We take all area vectors here to point exterior to the box. (a) Calculate the electric ¡ux through the vertical rectangular surface. This box side is ¡at, having area A = (0 : 30 m ) (0 : 10 m ) = 0 : 030 m 2 : Because the electric ¢eld is constant on this ¡at surface, we obtain Z E ¡ da = E ¡ A = EA cos 180 & = & 7 : 80 & 10 4 ¡ (0 : 030) ( ¢ 1) = ¢ 2340 N ¡ m 2 = C as the electric ¡ux through the vertical rectangular surface. (b) Calculate the electric ¡ux through the slanted surface. This box side is ¡at, having area A = (0 : 30 m ) ¢ : 10 m cos 60 & £ = 0 : 060 m 2 : Because the electric ¢eld is constant on this ¡at surface, we obtain Z E ¡ da = E ¡ A = EA cos 60 : & = & 7 : 80 & 10 4 ¡ (0 : 060) ¢ 1 2 £ = +2340 N ¡ m 2 = C as the electric ¡ux through the slanted rectangular surface. (c) Calculate the electric ¡ux through the entire surface of the box. The box£s remaining sides are all ¡at. Because the electric ¢eld is constant on each of these sides, we obtain Z E ¡ da = E ¡ A = EA cos 90 & = 0 for each remaining side : Therefore, we obtain I box E ¡ da = ¢ 2340 + 2340 + [ terms of zero ] = 0 as the electric ¡ux through the entire box surface. FYI: It follows from Gauss£s law for electricity...
View Full Document

Page1 / 3

workout 24 - HPHY 253 Workout 24 Gauss&s Law for...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online