Three point charges `+10nC,-15nC and +20nC` are placed at the corners A,B and C respectively of a square, ABCD of side 0.1m. Calculate the amount of work done to transfer -100nC from the point 'D' to the point 'O'.

Explanation

`V=((1)/(4piepsi_(0)))((q)/(x))` volt
`AC=(sqrt(2))a=sqrt(2)xx0.1m=0.14m`
`AO=BO=CO=DO=(0.14m)/(2)=0.07`
`V_(D)=9xx10^(9)xx10xx10^(-9)[(1)/(0-1)-(1.5)/(0.1sqrt(2))+(2)/(0.1)]`
`=90[30-10.605]`
`=1745.55V`
Similarly, `V_(@)=(9xx10^(9)xx10xx10^(-9))/((0.1//sqrt(2)))[1-1.5+2]`
`V_(0)=1909.19V`
`therefore p.d.=[V_(O)-V_(D)]=1909.19-1745.55`
Hence, `W=(pd)` (charge tranferred)
`=163.64xx(-100nC)=163.64xx-100xx10^(-9)`
`1.636xx10^(-5)J`.