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E+=(Ï/3ε) r1
E-=(Ï/3ε) r2 ....(Ï is not taken as negative as r2 is taken accordingly in opposite direction..)
ER=E++E-
(Ï/3ε)(r1+r2)
(Ï/3ε)r .. r is given unit vector joining two centers...
In a uniform sphere of charge density P (rho), a small cavity is created.
The center of the cavity is at a distance a from the center of the sphere. Taking center of the sphere as origin and vector a as unit vector along the line joining the center of cavity and origin find the electric field at a point inside the cavity.
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5 Answers
24its easy........apply superposition.....
E+=Ï/3ε r1
E-=Ï/3ε r2
ER=E++E-
i hope remaining 2 steps u can do..............
(just use r1+r2=a)
62but eureka.. will it give the electric field at any aribitrary point if that point is not specified?
62you have to consider the radius vector of that point as r
and then take the electric field due to both these seperately... using guass's law.
Or am I still into my last night's sleep :)
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