Question
A square surface of side $$L$$ metre is in the plane of the paper. A uniform electric field $$E\left( {\frac{V}{m}} \right),$$ 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
A square surface of side $$L$$ metre is in the plane of the paper. A uniform electric field $$E\left( {\frac{V}{m}} \right),$$ 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
A.
$$\frac{{E{L^2}}}{{\left( {2{\varepsilon _0}} \right)}}$$
B.
$$\frac{{E{L^2}}}{2}$$
C.
zero
D.
$$E{L^2}$$
Answer :
zero
Solution :
As we know, the electric flux $$\left( \phi \right)$$ through any surface area is given by,
$$\phi = E \cdot ds = \left| E \right|\left| {ds} \right|\cos \theta $$
As according to question, surface area is in plane of paper and $$E$$ is also in plane of paper. So, angle between area vector and $$E$$ is $${90^ \circ }$$
So, $$\phi = \left| E \right|\left| {ds} \right|\cos {90^ \circ } = {0^ \circ }$$
As we know, the electric flux $$\left( \phi \right)$$ through any surface area is given by,
$$\phi = E \cdot ds = \left| E \right|\left| {ds} \right|\cos \theta $$
As according to question, surface area is in plane of paper and $$E$$ is also in plane of paper. So, angle between area vector and $$E$$ is $${90^ \circ }$$
So, $$\phi = \left| E \right|\left| {ds} \right|\cos {90^ \circ } = {0^ \circ }$$
