# Consider a metal spher of radius ‘R’ that is cut in two parts along a plane whose minimum distance from the sphere’s centre is ‘h’. Sphere is uniformly charged by a total electric charge ‘Q’. The minimum force necessary to hold the two parts together, is?

Ranjan
13 Points
4 years ago
sphere = R                    surface charge density = Q / (4 pi R^2)
Curved surface
area of the smaller part S1 = 2 pi R (R - H) = 2 pi R^2 - 2 pi R H
Curved Surface
area of larger part S2  = 2 pi R ( R + H)
Both parts
have a plane area also = pi (R^2 – H^2).
On a metal the charges remain on the surface.  Let us say that the charges are distributed  on the total surface areas.  Because the sphere is being cut the charges
get redistributed.
Total
surface area of the two parts after cutting = S = 4 pi R^2 + 2 pi *(R^2 – H^2)
S =
6 pi R^2 – 2 pi H^2                  Surface charge density = Q/[2pi (3 R^2 –
H^2)]
Charge on
part S2  = [ pi (3R^2+2RH-H^2) ] * Q/[2
pi (3R^2-H^2)]
= Q
*[3 R^2+2RH-H^2]/[6R^2 -  2 H^2]
Charge on part
S1 = [3 R^2  - 2 RH – H^2]*Q/[6 R^2 – 2 H^2]
We need to
find the center of the charge distribution on the two spherical pieces.  We have to find center of masses of surfaces
ie., treating them as spherical shells of uniform thickness t.
For the small part S1:        COM = 1/(2 pi R(R-H))   integral   x * 2
pi R  dx   …. x varies from H  to R.
COM from
origin O = 1/(R-H) * (R^2 – H^2) /2   =
(R+H)/2
For the
large part S2:   COM = 1/ (2 pi R(R+H))  integral x 2 pi R dx …. X varies from –R to 0
and 0 to +H.
= COM = 1/ (R+H)  * (R^-H^2)/2
= (R-H)/2   to the left of  origin from origin.
Distance C1
C2 between the centers of charge distributions :  R
Force between
them:
1/(4 pi
e) * Q^2 [3 R^2+2RH-H^2] [3 R^2-2RH-H^2]/ {[6R^2 -  2 H^2]^2 *R^2 }
==
If we
consider that the charges distributed between two parts are in proportion of
the curved surface areas only, then:
Force =
1/(4 pi epsilon)  Q(R+H) /2R * Q(R-H)/2R  * 1/R^2
= 1/(4 pi epsilon) * Q^2 (R^2 – H^2) /(4R^4)