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# I need the answer to question no.8 .It is previous years jee advanced question.

Saurav
224 Points
3 years ago
BY GAUSS LAW
the surface electric field on the surface of outermost sphere will be
∫E.ds=(Q1+Q2+Q3)/$\large \varepsilon$0
THUS TOTAL CHARGE ON THE SURFACE WILL BE Q1+Q2+Q3
SIMILARLY THE CHARGE ON MIDDLE AND INNER SPHERES WILL BE Q1+Q2,Q1 RESPECTIVELY
BUT ALL OF THEM HAS GOT THE SAME CHARGE DENSITY
THEREFOR EQUATE THERE SURFACE CHARGE DENSITIES
YOU WILLL GET (Q1+Q2+Q3)/9 =(Q1+Q)/4 =Q1
THEN YOU WILL GET  THE RATIO   1:3:5