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At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere , the potentials are 100 V and 75 V respectively . Then which of the following statements are correct $\; $ 1. Potential at it's surface is 150 V $\;$2.The charge on the sphere is $\;\large\frac{50}{3}\;\times 10^{-10}\;C$ $\;$3.Electric field on the surface is $\;1500 V/m\;$ $\;$4.Electric potential it's centre is 25V $\;\quad \;$

$(a)\;1\;and\;2\qquad(b)\;1 , 2 \;and\;3\qquad(c)\;2\;and\;3 \qquad(d)\;2 , 3\;and\;4$

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Answer : (b) 1 , 2 and 3
Explanation :
Let the charge on the sphere be q and radius R . Then
$\large\frac{k\;q}{R+5\times10^{-2}}=100----(1)$
$\large\frac{k\;q}{R+10\times10^{-2}}=75----(2)$
$\large\frac{(1)}{(2)}\;we\;get\;\large\frac{R+10^{-1}}{R+5\times10^{-2}}=\large\frac{100}{75}=\large\frac{4}{3}$
$3R+3\times10^{-1}=4R+2\times10^{-1}$
$R=10^{-1}$
$R=10 \;cm$
Putting in equation (1) we get
$\large\frac{9\times10^{9}\times}{15\times10^{-2}}=100$
$q=\large\frac{15\times100\times10^{-11}}{9}$
$q=\large\frac{5}{3}\times10^{-9}\;C$
Potential on it's surface is
$V_{S}=\large\frac{k\;q}{R}=\large\frac{9\times10^{9}\times\large\frac{5}{3}\times10^{-9}}{10^{-1}}$
$V_{S}=150 V$
Electric field on it's surface
$E_{S}=\large\frac{k\;q}{R^2}=\large\frac{9\times10^{9}\times \large\frac{5}{3}\times10^{-9}}{10^{-2}}$
$E_{S}=1500 V/m$
Potential at centre of sphere
$V_{0}=\large\frac{k\;q}{2R^3}\;(3R^2-r^2)|_{r=0}$
$V_{0}=\large\frac{3\;k\;q}{2R}$
$V_{0}=\large\frac{3}{2} \times 150$
$V_{0}=225 V $
answered Feb 19, 2014 by yamini.v
 

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