Ncert Exercises & Solutions

Consider an electron in front of the metallic surface at a distance d (treated as an infinite plane surface). Assume the force of attraction by the plate is given as $\frac{1}{4} \frac{q^{2}}{4 π ε_{0} d^{2}}$ . 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]

Hint: The work done by the external force will be equal to the work done by the attractive force.

Step 1: Find the work done by the external force.

According to the question, consider the figure given below.

From figure, d=0.1 nm=10 $^{- 10}$ m,

$F = \frac{q^{2}}{4 \times 4 π ε_{0} d^{2}}$

Let the electron be at distance x from the metallic surface. Then, force of attraction on it is:

$F_{x} = \frac{q^{2}}{4 \times 4 π ε_{0} x^{2}}$

Work done by external agency in taking the electron from distance d to infinity is:

$\begin{aligned} W & = \int_{d}^{\infty} F_{x} d x = \int_{d}^{\infty} \frac{q^{2} d x}{4 \times 4 π ε_{0}} \frac{1}{x^{2}} \\ = \frac{q^{2}}{4 \times 4 π ε_{0}} [\frac{1}{d}] \\ = \frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times 10^{- 10}} J \\ = \frac{{(1.6 \times 10^{- 19})}^{2} \times (9 \times 10^{9})}{(4 \times 10^{- 10}) \times (1.6 \times 10^{- 19})} eV = 3.6 eV \end{aligned}$

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