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. `3 xx 10^(5) V`
B. `6xx 10^(5) V`
C. `9 xx 10^(5) V`
D. `18 mV`

Answer

A

Explanation

`V = (1)/(4pi epsilon_(0)) [(q)/(AC)+(q)/(BC)] = (q)/(4pi epsilon_(0))[(AC + BC)/(AC xx BC)]`
or `V = (9 xx 10^(9) xx 1 xx 10^(-6))/(1) xx [(10)/(30) xx 100] "volt"`
`= 3 xx 10^(5) "volt"`