Class Management I Help HW 2 Gauss\'s Law Begin Date: 1/27/2017 11:58:00 PM-Due
ID: 3161626 • Letter: C
Question
Class Management I Help HW 2 Gauss's Law Begin Date: 1/27/2017 11:58:00 PM-Due Date: 23/2017 7.00:00 PM End Date: 2/3/2017 7:00:00 PM (6%) Problem 4: An electric field E() Edarsi 2j+ 5k) passes through the cube with sides that are parallel to the x-y-zplanes with a side length of Lthat is centered on the origin (see figure). Otheexpertta.com P A 33% Part (a) Integrate to find the expression for the total electric flux through the cube. Grade Summary Potential 100% Submissions 4 5 6 Attempts remaining: 7 213 BACKSPACE tea CLEAR Submit I give up! Feedback: 0% deduction per feedback. Hints: 0% deduction per hint. Hints remaining S 33% Part (b) If the side L 6A m, a 1.1 m 3, and the field strength Eo 26 VIm what is the numerical value for the flux in v m? 33% Part (e) If the electric field was Ex Eoop i 2j+ 15k), what will be the total flux? All content 2017 Expert TA LLCExplanation / Answer
A] Flux = E(x).da
Now the flux through 4 sides is zero as 4 of them cancel out
Flux = E0*ax^3*L^2 + E0*ax^3*L^2 = E0*a*L^5/4
B] Flux = 26*1.1*6.4^5/4
= 76772 Vm
C] Flux = 0
as it get cancelled out