A charge 5C is placed at the centre of shell of radius, r = 3 m and having charges 5 C. The potential at a point `r/2` distance from the centre of the shell will be

Options

A. `-9 times 10^(9)V`
B. `30 times 10^(9)V`
C. `45 times 10^(9)V`
D. `-15 times 10^(9)V`

Answer

C

Explanation

Given, charge on the surface of the shell, q = 5C
radius of shell, r = 3 m
Now, the electric potential at a point at a distance r from the centre of the shell is given as,
`therefore " "V=1/(4piepsilon_(0))*q/r " ...(i)"`
`" "V_(1)=9 times 10^(9) times 5/3" "[because 1/(4piepsilon_(0))=k=9.0 times 10^(9)]`
`" "V_(1)=15 times 10^(9)V`
`therefore` Electric potential at r/2 due to charge at the centre, from Eqs. (i), we get
`" "V_(2)=9 times 10^(9) times 5/((3//2))" "({:(because r=3), (therefore r//2=3//2):}]`
`" "V_(2)=30 times 10^(9)V`
`therefore` Total potential,
`therefore " "V=V_(1)+V_(2)=(15 times 10^(9)+30 times 10^(9))`
`" "V=(15+30) times 10^(9)`
`" "V=45 times 10^(9)V`