Two point charges 2 `muC` and 0.1 `muC` are held at a separation of 20 mm. What work is required by the external agent to move the charges to half of the given separation?

Explanation

Here, Work done = Final potential energy - Initial potential energy
`rArr W=(1)/(4pi epsilon_(0)) (q_(1)q_(2))/(r.)-(1)/(4pi epsilon_(0)) (q_(1)q_(2))/(r )`
Here, `q_(1)=2muC= 2xx10^(-6)C`
`q_(2)=0.1muC=0.1 xx 10^(-6)C`
`r.=` Final separation between `q_(1) and q_(2)`
`=10 m m=10xx10^(-3)m`
r = Initial separation between `q_(1) and q_(2)`
`=20 mm=20xx10^(-3)m`
`therefore W=(q_(1)q_(2))/(4pi epsilon_(0)) [(1)/(r.)-(1)/(r )]`
`=9xx10^(9)xx2xx10^(-6)xx0.1xx10^(-6)`
`[(1)/(10xx10^(-3))-(1)/(20xx10^(-3))]`
`=9xx10^(9)xx2xx10^(-6)xx0.1xx10^(-6)xx(1,000)/(20)`
`=9xx10^(-2)J`