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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 dlt 0.1 nm]. |
Answer» Solution : 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 |
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