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Electric Field and Potential — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsElectric Field and Potential5 Marks
Question
A particle A having a charge of 2.0 × 10-6C is held fixed on a horizontal table. A second charged particle of mass 80g stays in equilibrium on the table at a distance of 10cm from the first charge. The coefficient of friction between the table and this second particle is $\mu=0.2.$ Find the range within which the charge of this second particle may lie.
✓

Answer


$\text{q}_\text{A}=2\times10^{-6}\text{C},\ \text{M}_\text{b}=80\text{g},\ \mu=0.2$

Since B is at equilibrium,

So, $\text{Fe}=\mu\text{R}$

$\Rightarrow\frac{\text{Kq}_\text{A}\text{q}_\text{B}}{\text{r}^2}=\mu\text{R}=\mu\text{m}\times\text{g}$

$\Rightarrow\frac{9\times10^9\times2\times10^{-6}\times\text{q}_\text{B}}{0.01}=0.2\times0.08\times9.8$

$\Rightarrow\text{q}_\text{B}=\frac{0.2\times0.08\times9.8\times0.01}{9\times10^9\times2\times10^{-6}}$

$=8.7\times10^{-8}\text{C}$

Range $=\pm8.7\times10^{-8}\text{C}$

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