Two positive charges of `10 muC and 5 muC` are 12 cm apart. Find the work done in bringing them 6 cm closer.

Explanation

Case-I : `q_(1)=10muC=10xx10^(-6)C`
`q_(2)=5muC=5xx10^(-6)C`
Distance between `q_(1) and q_(2)=12 cm`
`=12xx10^(-2)m`
`P.E.=U_(1)=(1)/(4pi epsi_(0))(q_(1)q_(2))/(r_(1))`
`=9xx10^(9)xx(10xx10^(-6)xx5xx10^(-6))/(12xx10^(-2))`
`=(9xx10xx5)/(12)xx10^(-1)`
`=(9xx5)/(12)=(45)/(12)=3*75J`
Case II : When the charges are brought 6 cm closer.
Then distance between the two charges `=r_(2)=12-6=6cm`
`r_(2)=6xx10^(-2)m`
`P.E.= U_(2)=(1)/(4pi epsi_(0))(q_(1)q_(2))/(r_(2))`
`U_(1)=9xx10^(9)xx(10xx10^(-6)xx5xx10^(-6))/(6xx10^(-2))`
`=(9xx10xx5)/(6)xx10^(-1)`
`=(45)/(6)=7*5J`
So work done to bring then 6 cm closer,
`W=U_(2)-U_(1)=7*5-3*75`
`=3*75J`