Question #22c70

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Answer 1 · 2017-04-07T08:15:49

Charge density increases in following ratio:

$sigma_2/sigma_1 = 2^(1/3)$

Initial volume:

$V_1 = 4/3 pi R_1^3 color(red)(times 2)$

Final Volume:

$V_2 = 4/3 pi R_2^3$

$= 4/3 pi R_1^3 times 2 implies R_2 = 2^(1/3) R_1$

Initial Surface Area:

$S_1 = 4 pi R_1^2 color(red)(times 2)$

Initial Charge density:

$sigma_1 = (sum Q)/(4 pi R_1^2 times 2)$

Final Surface Area:

$S_2 = 4 pi R_2^2 = 4 pi 2^(2/3) R_1^2$

Final Charge density:

$sigma_2 = (sum Q)/(4 pi 2^(2/3) R_1^2)$

$sigma_2/sigma_1 = ( (sum Q)/(4 pi 2^(2/3) R_1^2))/((sum Q)/(4 pi R_1^2 times 2)) = 2^(1/3)$