Question

(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement. (b) Two identical circular loops '1' and '2' of radius R each have linear charge densities −λ and +λ C/m respectively. The loops are placed coaxially with their centres R3 distance apart. Find the magnitude and direction of the net electric field at the centre of loop '1'.

Answer 1

In electromagnetism, electric flux is the measure of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a normally perpendicular surface. If the electric field is uniform, the electric flux passing through a surface of vector area S is \Phi_E = \mathbf{E} \cdot \mathbf{S} = ES \cos \theta, where E is the electric field (having units of V/m), E is its magnitude, S is the area of the surface, and θ is the angle between the electric field lines and the normal (perpendicular) to S. For a non-uniform electric field, the electric flux dΦE through a small surface area dS is given by d\Phi_E = \mathbf{E} \cdot d\mathbf{S} (the electric field, E, multiplied by the component of area parallel to the field). The electric flux over a surface S is therefore given by the surface integral: \Phi_E = \iint_S \mathbf{E} \cdot d\mathbf{S} where E is the electric field and dS is a differential area on the closed surface S with an outward facing surface normal defining its direction.