Question

A square frame of edge 10cm is placed with its positive normally making an angle 600 with a uniform electric field of 20v/m. Find the flux of the electric field through the surface bound by the electric frame

Answer 1

A square frame of edge 10cm is placed with its positive normally making an angle 600 with a uniform electric field of 20v/m. Find the flux of the electric field through the surface bound by the electric frame

solution : area of square frame of edge 10cm , A = (10cm)² = 100cm² = 100 × 10^-4 m² = 0.01 m²

electric field intensity, E = 20V/m.

angle between electric field intensity and normal to the area , α = 60°

we know, electric flux is the dot product of electric field intensity vector and area vector.

i.e., $\Phi=\vec{E}.\vec{A}$

= $|E|.|A|cos\alpha$

= 20V/m × 0.01 m² × cos60°

= 0.2 × 1/2

= 0.1 Vm

hence, electric flux through the surface of square frame is 0.1 Vm

Answer 2

Answer:

0.1vm

Explanation:

The surface considered is plane and the electric field is uniform (figure 30.). Hence, the flux is gtbrgtΔΦ=E→.ΔS→

=EΔScos600

+(20Vm−1)(0.01m2)(12)=0.1Vm