Question

A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and Gauss's Law, find the flux of the electric field due to this charge through the surface of the hemisphere (figure 30-E3).
Figure

Answer 1

Magnitude Q’s charge is present within the hemisphere, but not enclosed by a single hemisphere completely. So, the charge inside the sphere is Q. The total flux is distributed equally in two hemispheres by symmetry. Thus, through one hemisphere, the flux will be half of the flux via the full sphere.

Explanation:

According to Gauss law,

• For a spherical surface $F | u x=\oint \vec{E} \cdot d s=\frac{q}{\varepsilon_{0}}[b y \text { Gauss law }]$
• For a hemisphere, the total surface area is q/2ε0.  
• So, for a open bowl (hemisphere), the total flux via the curved surface is ϕ'=Q∈0$\times$12=Q2 ∈0.

Answer 2

Question:

A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and Gauss's Law, find the flux of the electric field due to this charge through the surface of the hemisphere (figure 30-E3).

Solution:

Let us complete the sphere by drawing remaining hemisphere dotted as shown in image attacted.

Then total electric flux through complete sphere = q /  ε₀ = Ψ (say)

So, flux through lower hemisphere = q/2ε₀ = Ψ/2