Question

A particle having a charge $10mC$ is held fixed on a horizontal surface. A block of mass $80g$ and having charge stays in equilibrium on the surface at a distance of $3cm$ 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.
$F_e=\mu mg$
$\frac{1}{4\pi {\epsilon }_0}\frac{q\times 10\times 10^{-6}}{{\left(3\times 10^{-2}\right)}^2}=0.5\times 80\times 10^{-3}\times 10$

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