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A spherical drop of water bas 3 xx 10^(10)C amount of charge residing on it. 500 V electric potential exists on its surface. Calculate the radius of this drop. If eight such drops (having identical charge and radii) combine to form a single drop, calculate the electric potential on the surface of the new drop. (k = 9 xx 10^(9) SI) |
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Answer» SOLUTION :Suppose R is the radius. The CHARGE and potential on the surface are `3 xx 10^(10)` C and 500 V. `:. V=(kQ)/(R)` `:.R=(kQ)/(V)` `= (9xx10^(9)xx3xx10^(-10))/(500)` `= 5.4 xx10^(-3)` m `= 0.54 xx10^(-2)` m `:. R = 0.54 ` cm The big drop is formed from combined 8 drops. The radius and volume of the big drop are R. and V.. `:. V = 8 V` `:. (4)/(3) pi (R)^(3) = 8XX(4)/(3) pi (R)^(3)` `:. (R)^(3) = 8(R)^(3) :. R = 2R` `:. R . = 2xx0.54= 1.08 cm = 1.08 xx10^(-2)`m The total charge on big drop `Q=8xx3xx10^(-10)= 24 xx10^(10)` C `:.` The electric potential of big drop `V = (kQ)/(R)= (9xx10^(9)xx24xx10^(10))/(1.08xx10^(-2))` `:. V = 2000 V ` Second Method : Suppose radius of drop is R . Charge on its surface is `3xx10^(10)` C. Hence potential on its surface is 500 V. `:. V = (kQ)/(R)` `:. R = (kQ)/(V) = (9xx10^(9)xx3xx10^(-10))/(500)` `:. R = (2.7)/(500) = 0.0054 ` m `:. R = 0.54 ` cm Potential on surface of new drop in given condition `V.=n^(2/(3))V` `= (8)^((2)/(3))xx500` `=4xx500` `:. V = 2000V` |
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