Two charges, each equal to `1muC`, are placed at the vertices `A` and `B` of a triangle `ABC`. The produced of `AC` and `BC` is `30 cm^(2)`. The sum of the sides `AC` and `BC` is `10 cm`. The potential at `C` is

Options

A. `3xx10^(5)V`
B. `6xx10^(5)V`
C. `9xx10^(5)V`
D. `18 mV`

Answer

1

Explanation

AC = x, x+y = 10
BC=y,xy = 30

`V_(c)=kq[(1)/(x)+(1)/(y)]`
`=kq[(x+y)/(xy)]=(9xx10^(9)xx10^(-6)xx3010^(-2))/(30xx10^(-4))`
`(3)/(10)xx10^10^(6)=3xx10^(5))`