Question

A charge of 10μC is placed at the centre of a hemisphere of radius R = 10 cm as shown. The electric flux through the hemisphere (in MKS units) is
 
A. ${6 × 10^5}$
B. ${12 × 10^{15}}$
C. ${9 × 10^6}$
D. ${6 × 10^8}$

Answer 1

Hint:As we know, the number of electric field lines (i,e. magnitude) that pass through or intersect a particular area/volume is electric flux. As we know the formula for electric flux passing through a sphere, so we can use it for the hemisphere also by dividing it by 2.


Formula used:
Electric flux through the hemisphere ${\phi=\frac{1}{2}\frac{q}{\varepsilon_0}}$


Complete answer:
As we know, according to the Gauss Theorem; The total electric flux in a closed surface is equal to the charge enclosed in that body divided by permittivity. The electric flux is directly proportional to the net charge applied to the volume of the closed body.
As we can see in fig. A, the electric lines in the body A are only once considered for the electric flux, not the outer ones.
The electric flux is denoted by the ϕ.
Given,
${q = 10μC = 10 × 10^{‒6}} C$
${ {ε_0} = 8.854 × 10^{‒12} {\dfrac{C^2}{N {m^2}}} }$
Now according to the question electric flux is passes through the hemisphere –
Now as we know,
Electric flux through the sphere ${[ \phi=\frac{q}{\varepsilon_0} ]}$
Electric flux through the hemisphere ${\phi=\frac{1}{2}\frac{q}{\varepsilon_0}}$
${\phi=\frac{10\times{10}^{-6}}{2\times 8.854\times{10}^{-12}}}$
${\phi=0.564\times{10}^{-6+12}}$
${\phi=0.6\times{10}^6}$
${\phi=6\times{10}^5}$
So, the electric flux through the hemisphere of radius R = 10 cm is ${6 × 10^5} $
Thus, Option (A) is correct


Note: According to Gauss's law of electricity, the net electric charge q enclosed by any closed surface has a direct relationship to the electric flux over that surface. By remembering the formulas for basic symmetric surfaces one can solve the questions for different other surfaces also.