Flux Through Cube The cube in Fig. 24-25 has edge lengths of 1.40 m and is orien

Question

Flux Through Cube The cube in Fig. 24-25 has edge lengths of 1.40 m and is oriented as shown with its bottom face in the x-y plane at z = 0. Find the net flux through the cube if the electric field is given by the following.

Flux Through Cube The cube in Fig. 24-25 has edge lengths of 1.40 m and is oriented as shown with its bottom face in the x-y plane at z = 0. Find the net flux through the cube if the electric field is given by the following. (c) In each case, how much charge is enclosed by the cube? (5.55 N/C + [2.85 N/CA?m]y) E vector = (4.20 N/C) J (b) E vector =(2.85 N/CA m)y

Explanation / Answer

electric flux=E.ds=q/e0

egde of cube=1.40 m

so ds=1.40*1.40

whether it is dx or dy it is same

value of electric field is given in both cases

so we can find q

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