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14 Mar 2008 07:15:57 IST

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Field inside cavity

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If you a solid non conducting spherical sphere of radius R with charge of magnitude q distributed uniformly on it. there is a spehrical cavity of raidus R/8 with its centre at a distance R/2 made in it. find the electirc field inside the cavity as a function of distance from the centre.

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Sai Ganesh Bandiatmakur

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Joined: 17 Feb 2007

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15 Mar 2008 07:38:11 IST

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Consider a pt P in the cavity where u want to find dirn+ magnitude of E-field
Let O be the centre of larger sphere, and "&" be the charge/unit volume for the whole sphere(it gets simple to rmr and use and derive using charge/unit volume) and A be the centre of cavity and
Let Vector OA=a, vectorAP=b and vector OP=r
U can solve it in a method like this:
let the situation be imagined as a full sphere(without any cavities of charge density & containing another sphere of charge density " - &" inside the sphere..hence u get the cavity as no net charge in that area.
1. Field due to first "total and complete sphere" :
use gauss law(lets first find the magnitude)
E(4pi*r²)=&*(4pi*r³/3)/

( i considered the solid sphere to be uniform..if not, u can integrate &*4pi*r²dr within limits from 0 to r)
magnitude of E₁(due to 1st sphere) is &*r/3

Similiarly, u can find magnitude due to the smaller sphere as E₂₌ &*b/3

vector E₁=E₁(unit vector along r which is also r/(magnitude of r))
so we get vector E₁=(&/3

)(vector r)
and we get vector E₂=(&/3

)(vector b)
net field is Vector E₁+VectorE₂= (&/3

)(vector r - vector b) =as we considered smaller sphere as negativly charged(if first is positively charged), field direction due to 1st sphere wud be in the direction of vector OA whereas field direction due to 2nd sphere wud be towards the centre i.e opposite to vector b..)
vector r - vector b = vector a ( from triangle law)
so magnitude and direction of E field at a pt. inside the cavity does not vary with space and has magnitude= (&/3

)(a)
I derived for general case, put a=R/2 and &= q/(4piR³/3), note that it does not depend on "R/8" The E-FIeld wud depend only on distance b/w the 2 centres and wud give the same ans. even if it is R/10000!

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