In the given electric field →E=[α(d+x)^i+E0^j]NC; where α=1NC hypothetical closed surface is taken as shown in figure. The total charged enclosed within the close surface is abcϵ0k,then k=___
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Solution
ϕABCD=−acdunit ϕCDEF=−bcE0unit
ϕABEF=bcE0+c∫a0(d+x)dy =+bcE0+acd+c∫a0xdy =+bcE0+acd+cab∫b0xdx [Sincexb+ya=1⇒dxb=−dya] =[+bcE0+acd+acb2]unit Using Gauss law ϕnet=qinϵ0⇒qin=abcϵ02