a metal sphere of radius R is uniformly charged by total charge Q. sphere is cut in two along plane whose minimum distance from centre is H.what is force necessary to hold two parts
Answers
Radius of 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 }
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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)
