The length of each side of a cubical closed surface is l. If charge q is situated on one of the vertices of the cube, then if the flux passing through shaded face of the cube is qx∈0. Find x:
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Solution
The cube has six surfaces . When a charge q at center of the cube , the flux through each surface is ϕ=q6ϵ0 When q is placed at one of the vertices of the cube, the charge inside the cube is q/4. now flux ϕ=(q/4)6ϵ0=q24ϵ0 thus, x=24