Question

Five equal point charges with $q=10nC$ are located at $x=2$ , $4$ , $5$ , $10$ and $20m$ . Taking the value of permittivity of free space ${ϵ}_0=\frac{1}{36\pi }\times 10^{-9}{C}^2{N}^{-1}{m}^{-2}$ and potential at infinity to be zero, calculate the potential (in $\text{V}$ ) at $x=0$ ?

Answer 1

Answer: 99

Given $q=10nC=10\times 10^{-9}C$
The potential at the point $x=0$ , will be the sum of potential produced by the charges placed at $x=2,4,5,10$ and $20m$
Potential,
$V=\frac{q}{4\pi {ϵ}_0\times r}$
Then, $V_1=\frac{q}{4\pi {ϵ}_0\times 2}$
$V_2=\frac{q}{4\pi {ϵ}_0\times 4}$
$V_3=\frac{q}{4\pi {ϵ}_0\times 5}$
$V_4=\frac{q}{4\pi {ϵ}_0\times 10}$
$V_5=\frac{q}{4\pi \times 20}$
The resultant potential,
$V=V_1+V_2+V_3+V_4+V_5$
$=\frac{q}{4\pi {ϵ}_0}\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}\right]$
$=\frac{q}{4\pi {ϵ}_0}\left[\frac{10+5+4+2+1}{20}\right]$
$=\frac{1}{4\pi {ϵ}_0}\times \frac{10\times 10^{-9}\times 22}{20}$
$=\frac{1\times 36\pi }{4\pi \times 10^{-9}}\times \frac{10\times 10^{-9}\times 22}{20}$
$=99V$