Question

ASAP please

5. Question 4.22 Given the electric flux density D= x2(x + y) +9(3x – 2y) 2 Determine the following: a. Pv by applying Eq. (4.26). b. The total charge enclosed in a cube 2 m on a side, located in the first octant with three of its sides coincident with the x-y-and z axes and one of its corners at the origin. c. The total charge Q in the cube, obtained by applying Eq. (4.29). (This can be done using 6 surface integrals, one for each face of the cube) Hints: (4.26) V.D= Pro (differential form of Gauss's law) $.ds= l. (4.29) (integral form of Gauss's law)

Answer 1

You just need to remember the expression for gradient function to solve this question. After solving we will get all answer 0, Because an electric field is not converging in nature so it will give del.D=0 also the cube we assume will contain no charge if we calculate by the gauss law given. If you like my answer kindly rate it thankyou
Solica V is a gradient fonction as Represent 7.5 = V. ( 12 luty) û + (3n-2y)9 | 2 (2(nty)) + f ( 3n-2y) oy 2 + (-2) = So this on (6) as electric v. is o field. het's do part (c) first we have a lube as follow!

we have to take Area Vector of earn Surface perpendicular to surface. the sin sufare ds can be seen as asoof eds, - dady y

dsi = audzý, ds₂ - dudz (-9) ds; = dudyh , dsy = dudy (-) dss = dydz ñ dso = dy dgt û ) 6 Dids - Q 3u floods, = f fe 3u-2y) andz -a?. a at this serfect. yra No No

Soody = 2 / - (3x-2y) du dz = (yu-suoja (3) = (pya - 3a)(a) at this point yao ssoods, = sso blage oo s soodsy = 1 beast cũ). (-3) = 0 lý ).(-3) = 0

ass = Is acuty) dy dz ( 2 (my+93) (3) 2 (nata?) (a) at this point n=a = 21 a2+ a?) (a) 2 - 6 a3 203 deo = S(-2cuty) dy dz - - 2 (na+az) (a) at this pointon = 0 = -a3

So Net 6 Dids -a -3a + 3a3 = - 2 a 3 + 2a3 So we can say a cube of this field have not charge inside will be o. if we apply fome thing part (2) we will get some Answer because Q is not depend of Side lepth = a. to Q in Question (b) and (C) is o