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At a distance of 5 cm and 10 cm outward ...

At a distance of `5 cm and 10 cm` outward from the surface of a uniformly charged solid sphere, the potentials are `100 V and 75 V`, repectively. Then.

A

potential at its surface is `150 V`

B

the charge on the sphere is `(5//3)xx10^(-10)C`

C

the electric field on the surface is `1500 V//m`

D

the electric potential at its centre is `225 V`

Text Solution

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The correct Answer is:
A, C, D

Potential at 5 cm from surface`=(KQ)/(R+5)=100`
Potential at `10` cm from surface
`=(KQ)/(R+10)=75 rArr R=10` cm.
`:.` Potential at surface `=(KQ)/(R)=(100xx15)/10=150 V`
Electric field on surface`=(KQ)/R^(2)=(100xx15Vxxcm)/(100 cm^(2))`
`=1500 V//m`
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