(A) 2q/ε₀
(B) q/8ε₀
(C) q/ε₀
(D) 6qa²/2ε₀
(D) 6qa²/2ε₀
Doubt by Mannat
Solution :
Here a charge at the corner of a cube is not symmetrically enclosed by the it. So we imagine 8 identical cubes arranged together so that the charge lies at the center of a larger cube.
Now, the total electric flux at the through the the entire large cube would be
Ф=q/ε₀ [By Gauss Theorem]
Ф=q/ε₀ [By Gauss Theorem]
Sicne the large cube consist of eight small cube. So the electric flux through each of the small cube would be
1/8 × q/ε₀
=q/8ε₀
1/8 × q/ε₀
=q/8ε₀
Hence, (B) q/8ε₀, would be the correct option.
Similar Question :
What is the flux through each side of the cube of side 'a' if a point charge of q is at one of its corners?
(A) 2q/ε₀Similar Question :
What is the flux through each side of the cube of side 'a' if a point charge of q is at one of its corners?
(B) q/8ε₀
(C) q/ε₀
(D) q/24ε₀
(D) q/24ε₀
Solution :
As per the above discussion, the flux through the each small cube would be q/8ε₀
Since the charge is lying at the corner / vertex of the cube i.e. it lies on the three faces of the cube. On these 3 faces, the electric field is along the plane / perpendicular to the area vector (i.e.θ=90°), so electric flux would be zero and no electric flux passes through them.
So we are left three opposite faces through which the electric field would be parallel to the area vector.
So the net flux will be divided equally along the three faces.
So the net flux will be divided equally along the three faces.
Hence, the flux through each face of the smaller cube would be 1/3×q/8ε₀
= q/24ε₀
= q/24ε₀
Hence, (D) q/24ε₀, would be the correct option.
