Answer
64.8k+ views
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:
![](https://www.vedantu.com/question-sets/494c82fe-b20a-4685-b7d0-2f70c90b28cd3676342393941810892.png)
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.
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:
![](https://www.vedantu.com/question-sets/494c82fe-b20a-4685-b7d0-2f70c90b28cd3676342393941810892.png)
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)