Consider an electron in front of metallic surface at a distance d(treated as an infinite plane surface).Assume the force of attraction by the plate is given as `(1)/(4)(q^(2))/(4piepsi_(0)d^(2))` .
Calculate work in taking the charge to an infinite distance from the plate.Taking d=0.1 nm,find the work done in electron volts.[Such a force law is not valid for d`lt` 0.1 nm].

Explanation

For the sake of simplicity,writing x in place d in the formula ,given in the statement,
`F=(1)/(4)((1)/(4piepsi_(0)))(q^(2))/(x^(2))`
Now ,as shown in the figure ,amount of work to be done by external force in order to displace electron by amount dx ,away from the surface is
dW=F.dx `cos0^(@)`
dW=F.dx
`therefore intdW=intF.dx`
`therefore W=underset(d)overset(oo)int(1)/(4)((1)/(4piepsi_(0)))(q^(2))/(x^(2))dx`
`=(1)/(4)((q^(2))/(4piepsi_(0))){-(1)/(x)}_(d)^(2)`
`=(1)/(4)((q^(2))/(4piepsi_(0))){-(1)/(oo)-(-(1)/(d))}`
`(1)/(4)((1)/(4piepsi_(0)))q^(2)xx(1)/(d)`
`=(1)/(4)xx9xx10^(9)xx(1.6xx10^(-19))^(2)xx(1)/(0.1xx10^(-9))`
`=5.76xx10^(-19)J`
`=(5.76xx10^(-19))/(1.6xx10^(-19))eV`
`therefore` W=3.6 eV