Question

A charged particle carrying electric charge of + 10$\mu$C is to rest on the horizontal surface of a table when another charged particle carrying a charge of +10$\mu$C is at a distance of 20 m from 10$\mu$C. Determine the magnitude and direction of the frictional force acting on the charged particle placed on the table.

• A. $2.25\times {10}^{-3}N$
• B. $90\times {10}^{-2}N$
• C. $9\times {10}^{-4}N$
• D. $18\times {10}^{-4}N$

Answer 1

The correct option is A $2.25\times {10}^{-3}N$

The electric force on each charge is $F=\frac{Q^2}{4\pi {ϵ}_o{r}^2}$

$⟹F=\frac{9\times {10}^9\times ({10}^{-5}{)}^2}{(20{)}^2}=2.25\times {10}^{-3}N$

The frictional force should be equal to electric force to keep the charge at rest and the direction of frictional force should be opposite to that of electric force.

Since both charges are positive, electric field will be away from each other and hence friction force will act along the line joining the charges and towards the other charge.

friction force$=f=2.25\times {10}^{-3}N$