Question

A particle having a charge $10 \,mC$ is held fixed on a horizontal surface. A block of mass $80 \,g$ and having charge stays in equilibrium on the surface at a distance of $3\, cm$ from the first charge. The coefficient of friction between the surface and the block is $\mu=0.5$. Find the range within which the charge on the block may lie

• A. $-4 \times 10^{-9} C$ to $4 \times 10^{-9} C$
• B. $-2 \times 10^{-9} C$ to $2 \times 10^{-9} C$
• C. $-4 \times 10^{-19} C$ to $4 \times 10^{-19} C$
• D. $-2 \times 10^{-19} C$ to $2 \times 10^{-19} C$

Answer 1

Answer: (A)

The block is in equilibrium because electric force is balanced by friction force. If $q$ is +ve, friction will act towards left and if $q$ is -ve, friction will act towards right. We know friction force develops according to need from 0 to $\mu N$. To get maximum value of charge that can be given to the second particle, we should equate repulsive force by maximum friction force. If $q$ is + ve.
$F_{e}=\mu m g$
$\frac{1}{4 \pi \varepsilon_{0}} \frac{q \times 10 \times 10^{-6}}{\left(3 \times 10^{-2}\right)^{2}}=0.5 \times 80 \times 10^{-3} \times 10$

$q=4 \times 10^{-9} C$
If $q$ is negative
$F_{e}=-\mu m g$
$q=-4 \times 10^{-9} C =$ maximum,$-$ ve charge
$\therefore $ Range of charge $=-4 \times 10^{-9} C$ to $4 \times 10^{-9} C$