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# Two charges $q$ and $- 4q$ are held at a separation $r$ on a frictionless surface. Another charge is kept in such a way that they do not move if released. The value of the third charge and its position from $- 4q$ is:A) $- 2q,2r$B) $- 4q,2r$C) $q,r$D) $4q,2r$

Last updated date: 25th Feb 2024
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Hint: Now, from the above problem, we know that two charges are placed at a distance $r$ from each other and it is also given that a third charge is also placed in such a way that the other two charges do not move if they are released. Now, we know that opposite charges attract each other. So, this third charge will be placed at a distance $x$ from the second charge. Now, by adding the force exerted by the third charge to first and second and equating it to zero, we will get our answer.

Formula used:
Force between the two charges placed at distance $d$ is $F = \dfrac{{K \cdot {q_1} \cdot {q_2}}}{{{d^2}}}$ . Where, $K$ is proportionality constant known as the Coulomb's law constant and ${q_1},{q_2}$ are the two charges.

Complete step by step solution:

Now, the third charge is $q'$and it is placed at a distance $x$ from the charge $- 4q$ .
Now, using the formula for force between the charge $q$ and $q'$ ,
We get,
$F = \dfrac{{K \cdot q \cdot q'}}{{{{\left( {r + x} \right)}^2}}}$ ,
Now, using the same formula for force between the charge $- 4q$ and $q'$ ,
We get,
$F = \dfrac{{K \cdot \left( { - 4q} \right) \cdot q'}}{{{{\left( x \right)}^2}}}$
Now, equating the adding the above two equations and equating them to 0 because we know that the system is in equilibrium.
We get,
$\dfrac{{K \cdot q \cdot q'}}{{{{\left( {r + x} \right)}^2}}} + \dfrac{{K \cdot \left( { - 4q} \right) \cdot q'}}{{{{\left( x \right)}^2}}} = 0...........\left( 1 \right)$
Now, simplifying the equation, we get,
$\dfrac{1}{{{{\left( {r + x} \right)}^2}}} + \dfrac{{ - 4}}{{{{\left( x \right)}^2}}} = 0 \\ {x^2} - 4{\left( {r + x} \right)^2} = 0 \\ {x^2} = 4{\left( {r + x} \right)^2} \\$
Now, taking square root on both sides, we get,
$x = 2r + 2x \\ x = - 2r \\$
Which means that the third charge is placed at a distance $2r$ from the charge $- 4q$ .
Now, taking the forces exerted on the second charge by first and third charge.
${F_{23}} = \dfrac{{K \cdot \left( { - 4q} \right) \cdot q'}}{{{{\left( { - 2r} \right)}^2}}}$
${F_{13}} = \dfrac{{K \cdot \left( { - 4q} \right) \cdot q}}{{{{\left( r \right)}^2}}}$
Now, subtracting the above equations and evaluating them to 0.
$\dfrac{{K \cdot \left( { - 4q} \right) \cdot q'}}{{{{\left( { - 2r} \right)}^2}}} - \dfrac{{K \cdot \left( { - 4q} \right) \cdot q}}{{{{\left( r \right)}^2}}} = 0$
Now, simplifying the above equation,
$\dfrac{{q'}}{{4{r^2}}} - \dfrac{q}{{{r^2}}} = 0 \\ q' = 4q \\$
Now, the value of the third charge is $4q$ .

Hence, the correct option is D.

Note: In the given problem the system of the three charges are in equilibrium. So, the resultant force exerted on the one charge by the other two is equal to zero. So, we have calculated the force on the second charge by the other two charges to calculate the value of the third charge.