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A square sheet of side 'a' is lying parallel to XY plane at z = a. The electric field in the region is \(\vec E=cz^2\hat k\). The electric flux through the sheet is :(a) a4c(b) \(\frac{1}{3}\)a3c(c) \(\frac{1}{3}\)a4c(d) 0 |
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Answer» Option : (a) a4c Given, \(\vec E\) = \(cz^2\hat k\) z = a \(\phi\) = ? \(\phi\) = \(\int \vec E.d\vec s\) ∵ ds = \(dxdy\,\hat k\) \(\phi\) = \(\int(cz^2)\hat k\,dx\,dy\,\hat k\) \(\phi\) = \(\int\limits_0^zcz^2 dx\,dy\) \(\phi\) = \(cz^2\)\(\int\limits_0^zdx\,\int\limits_0^zdy\) \(\phi\) = \(cz^2\)\([x]_0^z\) \([y]_0^z\) \(\phi\) = \(cz^2\)[z].[z] \(\phi\) = cz4 \(\phi\) = ca4 (∵ z = a given) ∴ Option : (a) is correct |
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