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.)$- 4 \times {10^{ - 12}}C$ to $4 \times {10^{ - 12}}C$
B.)$- 2C$ to $2 \times {10^2}C$
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

Hint: In the given problem the condition is given that the block having some charge stays in equilibrium with the charge which is being fixed on the same horizontal surface. This implies that the force in between them is in equilibrium, in other words the friction force and the electrostatic force which is acting in between them are of the same magnitude.

Step By Step Answer:
So, when the electrostatic force is balanced by the maximum friction force.
Maximum friction force=${f_k}$= $\mu mg$
Where $\mu =$coefficient of friction$= 0.5$
$m$=mass of the block=$80g = 0.08kg$
$g$=acceleration due to gravity=$9.8$ m/$s^2$
So now maximum friction force=$0.5 \times 0.08 \times 9.8 = 0.392$

So approximately, maximum friction force=${f_k} = 0.4$ $N$----equation (1)

Now the electrostatic force ${f_e} =$ $\dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
Where, $K =$electrostatic constant $= 9 \times {10^9}$ $Nm^2/C^2$

${q_{1 = }}$value of charge given$= 10mC = 10 \times {10^{ - 3}}C$
${q_{2 = }}$the charge on the block which we have to find.

$r =$ distance between the two charges $= 3cm = 3 \times {10^{ - 2}}m$

Now ${f_e} = \dfrac{{ = 9 \times {{10}^9} \times 10 \times {{10}^{ - 3}} \times {q_2}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}$------equation (2)

Now equating equation (1) and equation (2).

$0.4 = \dfrac{{ = 9 \times {{10}^9} \times 10 \times {{10}^{ - 3}} \times {q_2}}}{{{{(3 \times {{10}^{ - 2}})}^2}}}$
$\Rightarrow 3.6 = 9 \times {10^{ - 12}} \times {q_2}$
${q_2} = 4 \times {10^{ - 12}}C$

So, we have got the magnitude of the charge but here there are two possibilities that the charge may be positive or negative.

So, when the is positive then, ${q_2} = 4 \times {10^{ - 12}}C$
And when the ${q_2}$ is negative then, ${q_2} = - 4 \times {10^{ - 12}}C$
Hence the value of the charge will lie in between $4 \times {10^{ - 12}}C$ to $- 4 \times {10^{ - 12}}C$.

Hence the option (A) is the correct answer.

Note-The value of the friction is always in between 0 and $\mu mg$. So to calculate the maximum value of the charge which we can assign to the block to keep it in equilibrium with a fixed charge, already placed in the same plane we have to equate with the $\mu mg$ not 0, because this will give us the maximum charge.
Also, to develop the friction the value of the charge should be non-zero, otherwise the friction will not be developed.