Question

Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.06 x 104 N/C as shown in the figure below. 30.0 cm 60.0 10.0 cm (a) Calculate the electric flux through the vertical rectangular surface of the box. O kNm2/C O (b) Calculate the electric flux through the slanted surface of the box. kNm2/C (c) Calculate the electric flux through the entire surface of the box. kNm2/C Need Help? Read It Watch It

Answer 1

zom E = 7,06810* nlc . a= locm = loriölm 4 - 6.4 m b= 30 cm = 30x10m b= 0.30 m 64 a form " The electric fure through a surface is :: PE = Seids - Se dy code = E. S coso." de = E. ScoB -0 a) As the vertical rectangular surface is normal to the field. o=0; cose=1 Fix la : s=laxb) m2 07 DE = Exaxb) COS (0) - Ex caxb) collo) 3 - 7.06x004 x( 0.1 XO. 3) (1) I paz-2118 Naye lave for flux I going into the box) bi= -2.118 x103 Nmili 1$ = -2.118] kn.ml - -

. (b) slanted surface . 0 = 60 from triangle: since) 990.im 160 The (9.-) 50.2007 . sind . sin(30) ESCOS EX(bXAB) COSA s= (by MB) = (0.80x0.30) m? Pichant = 7.06 4104 x(0-2040.30) x cos 66 Poland' = 2118 Nm te 2,118 xr03 Nmille : : De = 2.183 % ku melc - © Sure thorough Entre' surtave : - Der tills + Pends- © E.S COSE ; where =90 i d ends = .. bendy = Es cosCao) = 0 Nmile fom -) 0 0.0 tot = (–2.1184103 )+ (2.118 %) + 0 Mitol 0.]knmye .- -> The entire surface of the box is a closed and as there no charge with in it, hence according to Gouys law the flux through the entire closed sunfaice is 2.20010)