Question

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 ${\displaystyle \frac{1}{4}\frac{q^2}{4\pi {\epsilon }_0{d}^2}}$. 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 < 0.1nm].

Answer 1

As per question F = ${\displaystyle \frac{1}{4}\frac{q^2}{4\pi {\epsilon }_0{d}^2}}$

Let at any instant electron is at a distance x from the metal surface. Force of attraction between. metal surface and electron is F

F = ${\displaystyle \frac{1}{4}\frac{q^2}{4\pi {\epsilon }_0{x}^2}}$

Work done by an external agency in taking the electron from distance d to ∞ is W.D. = ${\displaystyle {\int }_d^{\infty }F\cdot dx}$

W.D. = ${\displaystyle {\int }_d^{\infty }\frac{q^{2}dx}{4\times 4\pi {\epsilon }_0{x}^2}=\frac{q^2}{4\times 4\pi {\epsilon }_0}{\int }_d^{\infty }{x}^{-2}dx=\frac{q^2}{4\times 4\pi {\epsilon }_0}\frac{x^{-1}}{-1}}$

W.D. = ${\displaystyle \frac{-q^2}{4.4\pi g{e}_0}{\left[\frac{1}{x}\right]}_d^{\infty }=\frac{-q^{2}k}{4}[\frac{1}{\infty }-\frac{1}{d}]}$

W.D. = ${\displaystyle \frac{+k{q}^2}{4d}}$ [d = 0.1 nm = 10^–10 m]

Work done is positive

So W.D. = ${\displaystyle \frac{{(1.6\times {10}^{-19})}^2\times 9\times {10}^9}{4\times {10}^{-10}}}$

J = ${\displaystyle \frac{1.6\times 9\times 1.6\times {10}^{-33+9+10}}{4\times 1.6\times {10}^{-19}}}$ eV

= ${\displaystyle \frac{9\times 1.6\times {10}^{19+19}}{4}}$

= 3.6 eV