Assignment 2 Solutions Chpt 232. We use AdErr⋅=Φ∫and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m. (a) On the top face of the cube y= 2.0 m and ()ˆjdAdA=r. Therefore, we have ()()2ˆˆ4i32.02 j4i18jE=−+=−rˆˆ. Thus the flux is ()()()()222toptoptopˆˆˆ4i18jj18182.0N mC72 N mC.E dAdAdAΦ =⋅=−⋅= −= −⋅= −⋅∫∫∫rr(b) On the bottom face of the cube y= 0 and )ˆ)((jdAAd−=r. Therefore, we have jijiEˆ6ˆ4ˆ)20(3ˆ42−=+−=r. Thus, the flux is ()()()()222bottombottombottomˆˆˆ4i6jj662.0NmC24 NmC.E dAdAdAΦ =⋅=−⋅−==⋅= +⋅∫∫∫rr(c) On the left face of the cube ()()ˆidAdA=−r. So ()()()()222leftleftbottom ˆˆˆˆ4iji44 2.0N mC16 N mC.yE dAEdAdAΦ =⋅=+⋅−= −= −⋅= −⋅∫∫∫r(d) On the back face of the cube ()()ˆkdAdA=−r. But since Erhas no zcomponent . Thus, Φ= 0. 0E dA⋅=rr(e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 N·m2/C. Thus the net flux through the cube is Φ= (–72 + 24 – 16 + 0 + 0 + 16) N·m2/C = – 48 N·m2/C. 20. Using Eq. 23-11, the surface charge density is ()()5122202.310 N C8.8510C/ Nm2.010C/m .Eσε−−==××⋅=×6227. We assume the charge density of both the conducting cylinder and the shell are uniform, and we neglect fringing effect. Symmetry can be used to show that the electric 1
has intentionally blurred sections.
Sign up to view the full version.