Question

Two uncharged metal balls of radius 0.09mm each collide such that an electron is transferred between them. The potential difference between the balls is

Answer 1

Two uncharged metal balls of radius r = 0.09 mm each collide such that an electron is transferred between them. means, charge on first metal ball is +e then charge on other metal ball is -e , where e denotes magnitude of charge on electron.

Now from formula of potential,

potential of first metal ball, $V_1=\frac{ke}{r}$

potential of another metal ball, $V_2=-\frac{ke}{r}$

so, potential difference, $\Delta V=V_1-V_2=2\frac{ke}{r}$

here, k = 9 × 10^9 Nm²/C², e = 1.6 × 10^-19C and r = 0.09 mm = 9 × 10^-5 m

so, ∆V = 2 × 9 × 10^9 × 1.6 × 10^-19/(9 × 10^-5)

= 3.2 × 10^-5 Volts

= 32 $\mu V$

hence, answer is 32 $\mu V$