###### Ranjan

Last Activity: 5 Years ago

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 }

==

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)