A point charge +Q is place just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct ?

Options

A. Total flux through the curved and the flat surface is `Q/epsilon_0`
B. The component of the electric field normal to the flat surface is constant over the surface
C. The circumference of the flat surface in an equipotential
D. The electric flux party through the curved surface of the hemisphere is `-Q/(2epsilon_0) (1-1/sqrt2)`

Answer

C, D

Explanation

Flux through a closed three dimensional surface is `Q//epsilon_0`, where Q is enclosed charged within the 3D surface. Here charge is outside the given surface hence we need to extend the surface to enclose the charge. Hence flux through given surface has to be less than `Q//epsilon_0`. Hence option (a) is wrong.
(b) Refer following figure.

`E bot =E cos theta`
When we move away from the centre, on flat circular face of hemisphere, then electric field intensity decreases due to increase in distance of the point from point charge. At the same time angle between electric field intensity and normal also increases. Hence component of electric field normal to the surface cannot be same for all points on circular face. Hence option (b) is wrong.
(c )Distance of any point on circumference of circular face from point charge is `Rsqrt2`. Hence there will be same potential for all points.
`V=1/(4piepsilon_0) Q/(sqrt2R)`
Hence option (c) is correct .
(d)Electric flux linked with curved surface is same as that with circular face of hemisphere, except that it is negative for curved surface whereas positive for the circular face. We can calculate flux for the circular face and then apply negative sign to write the flux for curved surface. Note that electric lines will enter the curved surface and the same will come out from flat circular face.
Solid angle made by circular face at the location of point charge can be written as follows:
`Omega=2pi(1-cos theta)`
Here `theta=45^@`
Total solid angse around the charge is `4pi` hence flux associated with circular face can be written as follows (Proportional to solid angle)
`phi=Omega/(4pi)xxQ/epsilon_0=-(2pi(1-cos theta))/(4pi) Q/epsilon_0`
`=Q/(2epsilon_0) (1-1/sqrt2)`
Same negative flux is associated with curved surface hence option (d) is correct.