Seems quite strange that i cant solve simple questions like this... but on the other hand applying Gauss Law to cylinders spheres and such isnt that hard for me... ANYWAY(adsbygoogle = window.adsbygoogle || []).push({});

One of the cube's edges is placed at the origin.

A cube with edges 1.4m is oriented as shown in the diagram. Find the flux and the charge inside the cube when the electric field is

a) (3 Nm/c) yj

b) (-4N/C)i+ [6N/C + (3Nm/C)y]j

for a) since there are TWO faces in the X only. THe Y and Z componenets are orthogonal to the field thus are zero. [tex] \Phi = 2 \int E \bullet dA = \frac{q_{enc}}{\epsilon_{0}}[/tex]

[tex] \Phi = 2 \int 3y [d(y^2)] = 2 \int 6y^2 dy = 4y^3 [/tex]

since y = 1.4m flux is 10.976 Nm^2/C

the charge in teh cube as a result is [tex] 10.976=\frac{q_{enc}}{\epsilon_{0}} [/tex]

for b), since there are two faces for the X and Y sides

in the case of the X the field is constnat so [itex]\int E dA = EA[/itex]

[tex] \Phi = 12(1.4)^2 + 2 \int (6+3y)2y dy [/tex]

blah blah blah leads to

[tex] \Phi = 12(1.4)^2 + 2 [6(1.4)^2 + 2(1.4)^3] [/tex]

and plugging that equal to Qenc/epsilon yeilds the induced charge

Am i right? If i am wrong please do tell! Thank you so much!

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# Electric field Q

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