electric field

Question

What is the electric field at the centre of hemisphere of radius r and surface charge density as k C/m²

Answer 1

Hi Samagra!

 

By center of the hemisphere, do you mean the center of the base (aka the center of the sphere of which the hemisphere is half)?  I hope so, because if you mean the centroid of the hemisphere, the problem gets more mathematically intricate without necessarily adding any new physical concept.

 

So I will assume you mean the center of the base.  With that assumption, the procedure is essentially the same as for determining the electric potential at the center of the hemisphere, which was part of a previous problem you posted.  In this case, though, the charge is only on the surface of the hemisphere.  We have to figure out the contribution to the electric field at the center of the base of the hemisphere from each infinitesimal charge unit on the hemisphere surface, and then integrate those contributions over the entire hemisphere surface (using symmetry arguments where we can to simplify the process).

 

The only other wrinkle is that, since electric field is a vector, we have to pay attention to direction.

 

First of all, we have infinitesimal area units of of charge dq = k(r²sinφdθdφ).  This, again, assumes the angles are measured in radians (this expression is not true otherwise).  Note that dr is not involved here, because there is no thickness to this surface.  It is two dimensional.

 

Each infinitesimal area unit of charge in the sphere is a distance r from the center, since every point is on the surface of the spherical shell of radius r.

 

Hence, each infinitesimal area unit of charge on the hemisphere surface generates an electric field at the center with magnitude:

 

E = kdq/r² = Kk(r²sinφdθdφ)/r² = Kk(sinφdrdθdφ)  (where K is the Coulomg constant, to distinguish from k, the surface charge density)

 

The direction of each electric field contribution from each spot on the surface points radially away from the surface.  In a drawing, you might be able to see that, for every point on the hemisphere, there is another point located symmetrically around the sphere's symmetry axis, whose electric field would have equal magnitude, but would point in a mirror image fashion on the other side of the symmetry axis.  Once all those contributions are added up, the perpendicular-to-the-axis component of the electric field from one point on the hemisphere would always be cancelled out in the integral summation by the perpendicular component from the symmetric position on the hemisphere.  Therefore, the only component of the electric field from each point on the hemisphere at the center base that survives is the one that points directly along the symmetry axis (down, away from the hemisphere)

 

The magnitude of this electric field component, E_z, in polar coordinates terms, is given by

 

E_z = Ecosφ  (if φ is measured, as it conventionally is, from the +z axis, and the +z axis is the symmetry axis of the hemisphere) = Kk(sinφcosφdθdφ)

 

So, to get the total electric field, we have to integrate this expression over the entire hemisphere surface.  If the symmetry axis of the hemisphere is the +z-axis, with the base of the hemisphere at z=0, then the limits of integration to cover the entire hemisphere surface are 0 to 2π in θ and 0 to π/2 in φ.

 

The integral to get the resultant electric becomes:

 

E_tot = ∫₀^π/2∫₀^2π (Kksinφcosφdθdφ) = (Kk)∫₀^π/2∫₀^2π sinφcosφdθdφ  (taking all the constants outside the integral)

 

The θ integral here is pretty straightforward.  The φ integral may be simplified by realizing that sinφcosφ = (1/2)sin(2φ).  Integrating the latter expression (and using the chain rule appropriately) may make things easier.

 

I apologize that a lot of this would make more sense with a drawing, but I hope it can put you on the right track.  Just let me know if you want to go into any part of this further.