A charged sphere of diameter 4 cm has a charge density of $10^{- 4} C / {c m}^{2}$ . The work done in joules when a charge of 40 nano-coulombs is moved from infinity to a point, which is at a distance of 2cm from the surface of the sphere is:

14.4 π

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28.8 π

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144 π

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288 π

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Solution

The correct option is A $14.4 π$
Work done = Change in P.E.
Change in P.E. $Δ U = \frac{K Q q}{R + r} - 0$
where $K = 9 \times 10^{9}$ ,
$Q = σ 4 π r^{2} = 16 π \times 10^{- 4} C$
$q = 40 \times 10^{- 9} C$ ,
Putting all the values we get,
$W = \frac{9 \times 10^{9} \times 16 π \times 10^{- 4} \times 40 \times 10^{- 9}}{4 \times 10^{- 2}}$
$W = 14.4 π J$

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A charged sphere of diameter 4 cm has a charge density of 10−4C/cm2. The work done in joules when a charge of 40 nano-coulombs is moved from infinity to a point, which is at a distance of 2cm from the surface of the sphere is:

The correct option is A 14.4πWork done = Change in P.E.Change in P.E. ΔU=KQqR+r−0where K=9×109, Q=σ4πr2=16π×10−4C q=40×10−9C,Putting all the values we get, W=9×109×16π×10−4×40×10−94×10−2W=14.4π J

A charged sphere of diameter 4 cm has a charge density of $10^{- 4} C / {c m}^{2}$ . The work done in joules when a charge of 40 nano-coulombs is moved from infinity to a point, which is at a distance of 2cm from the surface of the sphere is: