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A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec{E}$ (volt $/ \mathrm{m})$, also in the plane of the paper, is limited only to the lower half of the square surface (see figure). The electric flux in SI units associated with the surface is:
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The correct answer is:
zero
Electric flux $\phi_E=\int \vec{E} \cdot \overrightarrow{d s}=\int E d s \cos \theta$
$$
=\int E d s \cos 90^{\circ}=0 .
$$
$$
=\int E d s \cos 90^{\circ}=0 .
$$
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