An electric field E→=4xi^−y2+1j^N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕ1 and ϕ2 respectively .The difference between ϕ1−ϕ2 is in Nm2/C ____
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answer is -0048.00.
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Detailed Solution
E¯=4xi^−y2+1j^.N/cArea of ABCD is dsABCD=lb k^ =3×2k^ =6k^ ϕ1=E¯.ds¯ABCD =0Area of BCGF dsBCGF=lb i^ =2×2i^ =4i^ ϕ2=E¯.ds¯BCGF =43i^−y2+1j^.4i^ ϕ2=12×4 =48Nm2/C ⇒ϕ1−ϕ2=−48.00Nm2/C