Question #22c70

Question

1720 views around the world

You can reuse this answer
Creative Commons License

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)#