A spherical hollow conductor is charged to -250V. If the radius of the conductor is 0.15m, then calculate
(i) Charge on the conductor.
(ii) Electric files intensity on the surface
(iii) Electric field intensity at the centre.
(iv) Electric potential difference between A and B at distances of 0.20 m & 0.25m from the centre respectively.

Explanation

(i) Using the formula `V=((1)/(4piepsi_(0)))((Q)/(R))`
`-250=(9xx10^(9)Q)/(0.15)`
`thereforeQ=(-250xx0.15)/(9xx10^(9))`
`Q=-4.166xx10^(-9)C`

(ii) `E=((1)/(4piepsi_(0)))(Q)/(R^(2))=(9xx19^(9)xx-4.166xx10^(-9))/((0.15)^(2))=1.666xx10^(3)Vm^(-1)`
(iii) Electric field intensity at any point inside a charged hollow spherical conductor is zero.
(iv) `V_(A)-V_(B)=(-Q)/(4piepsi_(0)))[(1)/(OA)-(1)/(OB)]=-9xx10^(9)xx4.166xx10^(-9)[(1)/(0.20)-(1)/(0.25)]`
`=37.494[(0.25-0.2)/(0.05)]=37.494[(25-20)/(500)]=(37.494xx100xx1)/(100)=37.494V`