Question #22c70

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Question #22c70

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Answer 1 · 2017-04-07T08:15:49

Charge density increases in following ratio:

$\frac{σ_{2}}{σ_{1}} = 2^{\frac{1}{3}}$ $sigma_2/sigma_1 = 2^(1/3)$

Explanation:

Initial volume:

$V_{1} = \frac{4}{3} π R_{1}^{3} \times 2$ $V_1 = 4/3 pi R_1^3 color(red)(times 2)$

Final Volume:

$V_{2} = \frac{4}{3} π R_{2}^{3}$ $V_2 = 4/3 pi R_2^3$

$= \frac{4}{3} π R_{1}^{3} \times 2 \Rightarrow R_{2} = 2^{\frac{1}{3}} R_{1}$ $= 4/3 pi R_1^3 times 2 implies R_2 = 2^(1/3) R_1$

Initial Surface Area:

$S_{1} = 4 π R_{1}^{2} \times 2$ $S_1 = 4 pi R_1^2 color(red)(times 2)$

Initial Charge density:

$σ_{1} = \frac{\sum Q}{4 π R_{1}^{2} \times 2}$ $sigma_1 = (sum Q)/(4 pi R_1^2 times 2)$

Final Surface Area:

$S_{2} = 4 π R_{2}^{2} = 4 π 2^{\frac{2}{3}} R_{1}^{2}$ $S_2 = 4 pi R_2^2 = 4 pi 2^(2/3) R_1^2$

Final Charge density:

$σ_{2} = \frac{\sum Q}{4 π 2^{\frac{2}{3}} R_{1}^{2}}$ $sigma_2 = (sum Q)/(4 pi 2^(2/3) R_1^2)$

$\frac{σ_{2}}{σ_{1}} = \frac{\frac{\sum Q}{4 π 2^{\frac{2}{3}} R_{1}^{2}}}{\frac{\sum Q}{4 π R_{1}^{2} \times 2}} = 2^{\frac{1}{3}}$ $sigma_2/sigma_1 = ( (sum Q)/(4 pi 2^(2/3) R_1^2))/((sum Q)/(4 pi R_1^2 times 2)) = 2^(1/3)$

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Question #22c70

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