A point charge q is placed at the centre of spherical shell what is flux through hemispherical surface
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Although the question is ambiguous, let me adress 2 of the most probable cases:
Case 1: the point charge is eaxctly at the geometric centre of the sphere of which the hemisphere is a part:
The flux through the flat part is zero. Note that gauss law can’t be applied here for charges on the boundary surface of the system. But intuitively its obvious that the field emanating from the point charge is spherically symmetric. Now the interesting part. This is a mathematical property of planes: any infinite plane divides a space 2 disjoint sets. Here, we dont need an infinite plane. But any radially emanating field line either :
1)goes above a plane passing through the charge
2)goes below the plane passing through the charge
3) moves along the plane passing through the charge.
Note that in none of the 3 cases, a field line pierces the surface, so the flux is zero.
Case 2: the point charge is slightly(ininitesimally) above the geometric center:
In this case, again consider the flat surface of the hemisphere to be a plane. This time, no. 2 of the previous list above passes through this surface. All others pass through the curved surface. So the flux is non-zero. Since the the charge is only infinitesimally displaced, no other change in flux through the curved surface occurs. This means that the flux is given by q/2ϵ0.q/2ϵ0. Half of the total flux.
Case 1: the point charge is eaxctly at the geometric centre of the sphere of which the hemisphere is a part:
The flux through the flat part is zero. Note that gauss law can’t be applied here for charges on the boundary surface of the system. But intuitively its obvious that the field emanating from the point charge is spherically symmetric. Now the interesting part. This is a mathematical property of planes: any infinite plane divides a space 2 disjoint sets. Here, we dont need an infinite plane. But any radially emanating field line either :
1)goes above a plane passing through the charge
2)goes below the plane passing through the charge
3) moves along the plane passing through the charge.
Note that in none of the 3 cases, a field line pierces the surface, so the flux is zero.
Case 2: the point charge is slightly(ininitesimally) above the geometric center:
In this case, again consider the flat surface of the hemisphere to be a plane. This time, no. 2 of the previous list above passes through this surface. All others pass through the curved surface. So the flux is non-zero. Since the the charge is only infinitesimally displaced, no other change in flux through the curved surface occurs. This means that the flux is given by q/2ϵ0.q/2ϵ0. Half of the total flux.
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