Question

If the surface normal vector makes an angle θ with the electric field E, then the electric flux through a surface of area dS is given by
(A) \[\phi = EdS\sin \theta \]
(B) \[\phi = EdS\cos \theta \]
(C) \[\phi = dS\cos \theta \]
(D) \[\phi = E\cos \theta \]

Answer 1

Hint: The scalar product or the dot product between the two vectors namely the normal vector to the surface and the electric field vector to be used.

Complete step by step solution:
We know that, electric flux is given by the formula,
\[\phi = \vec E.d\vec S\]
Where,
 \[\phi \] is the electric flux
 \[E\] is the electric field
  \[dS\] area of the surface
The direction of the area vector (\[d\vec S\]) is along the surface normal vector (\[\hat n\])
Thus, \[\phi = EdS\cos \theta \]

Thus the correct option is B.

Note: The dot product of the two vectors is generally the influence of the one vector over the other, which is the physical significance of this operation. This depends on the angle between the two and the cosine of the angle between them would determine the strength of this influence. This implies that if the two vectors are collinear, then their influence is greatest, while if they are perpendicular the effect would be zero. For the surface typically have varying curvature and each segment of the surface will have a different orientation and hence the perpendicular vector on each segment of the surface would be different. So when such a surface is subjected to the electric field, its influence would be dependent on the local angle between the electric field and the perpendicular vector (normal vector) at each point on this surface given by the cosine of angle between the two.