Question

Two particles each having a mass of $5g$ and charge ${10}^{-7}C$, stay in limiting equilibrium on a horizontal table, with a separation of $10cm$ between them. The coefficient of friction between each particle and the table is same. Find the value of this coefficient.

• A. $0.42$
• B. $0.18$
• C. $0.25$
• D. $0.12$

Answer 1

The correct option is B $0.18$

The particles are in equilibrium, which means net force on them is zero.

Thus, repulsion due to their charge = friction due to table. Since equilibrium is limiting, friction will attain its maximum value.

$\therefore \frac{k{Q}^2}{r^2}=\mu mg$

where $k=9\times {10}^{9}N{m}^2{C}^{-2}$ is the electrostatic constant, $Q={10}^{-7}C$ is the charge, and $r=0.1m$ is the distance between the charges.

$\frac{(9\times {10}^9)({10}^{-7}{)}^2}{(0.1{)}^2}=\mu \times (5\times {10}^{-3})\left(10\right)$

$\therefore \mu =0.18$

$\to QED.$