A charged sphere of diameter `4` cm has a charge density of `10^(-4) "coulmb"//cm^(2)`. The work done in joule when a charge of 40 nano-columbs is moved from infinity to a point which is at a distance of 2 cm from the surface of the sphere is-

Options

A. `14.4 pi`
B. `28.8 pi`
C. `144 pi`
D. `288 pi`

Answer

A

Explanation

`w= qDeltaV`
`sqrt(AB) = sqrt(2) = 1.4`
and error in `sqrt(AB)`
Let `Z = sqrt(AB)`
`(DeltaZ)/(Z) = (1)/(2) (DeltaA)/(A) + 1/2 (DeltaB)/(B)`
`(DeltaZ)/(1.4) = 1/2 ((0.2)/(1)) + 1/2 ((0.2)/(2))`
`(DeltaZ)/(1.4) = 0.1+ 0.05`
`Delta Z = (0.15)(1.4) = 0.21`
`Z = (1.4+0.2) m`