Answer
Verified
400.8k+ views
Hint: The state of equilibrium of the third charge will depend on the net force on it due to the other charges. As long as the net force remains zero, the charge will remain in the stable equilibrium and the charges will remain stationary.
Formula used:
The electrostatic force between two charges ${q_1}$ and ${q_2}$ is given by the Coulomb’s law by the following expression:
$F = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here r signifies the distance of charge from the point of observation and value of $\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
Detailed step by step solution:
At first, we have the three charges in the following configuration. We notice that the charges +q and +q are at equal distances from the charge –q. Using the expression for Coulomb law, we see that the force on charge –q due to the other two charges is equal and opposite. The net force can be written as follows using the expression for Coulomb’s law for force between two charges.
$\overrightarrow {{F_{net}}} = \overrightarrow F + \overrightarrow F = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q^2}}}{{{a^2}}} - \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q^2}}}{{{a^2}}} = 0$
Now if we start displacing the charge –q along the y-axis, we notice that the distance from the other two charges will increase but the magnitude of their forces will remain the same as we can see in the following diagram.
The forces acting on the charge –q will be resolved into horizontal and vertical components. Since the two positive charges have the same magnitude they will exert the same force on –q. Due to which, the horizontal components will cancel each other and while the vertical components will add up and the net force on –q will be given as
${F_{net}} = 2F\cos \theta $
We see that the force is not zero hence, there will be motion of charge –q towards the origin back to the stable equilibrium. Therefore the correct answer is option A because displacing the charge along the y-axis does not bring instability to the system.
Note: The student should note that when the charge –q is moving at some angle with respect to the other charges then the net force on the charge decreases with the increase in angle. Only the vertical component contributes to the force on the charge.
Formula used:
The electrostatic force between two charges ${q_1}$ and ${q_2}$ is given by the Coulomb’s law by the following expression:
$F = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here r signifies the distance of charge from the point of observation and value of $\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$
Detailed step by step solution:
At first, we have the three charges in the following configuration. We notice that the charges +q and +q are at equal distances from the charge –q. Using the expression for Coulomb law, we see that the force on charge –q due to the other two charges is equal and opposite. The net force can be written as follows using the expression for Coulomb’s law for force between two charges.
$\overrightarrow {{F_{net}}} = \overrightarrow F + \overrightarrow F = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q^2}}}{{{a^2}}} - \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q^2}}}{{{a^2}}} = 0$
Now if we start displacing the charge –q along the y-axis, we notice that the distance from the other two charges will increase but the magnitude of their forces will remain the same as we can see in the following diagram.
The forces acting on the charge –q will be resolved into horizontal and vertical components. Since the two positive charges have the same magnitude they will exert the same force on –q. Due to which, the horizontal components will cancel each other and while the vertical components will add up and the net force on –q will be given as
${F_{net}} = 2F\cos \theta $
We see that the force is not zero hence, there will be motion of charge –q towards the origin back to the stable equilibrium. Therefore the correct answer is option A because displacing the charge along the y-axis does not bring instability to the system.
Note: The student should note that when the charge –q is moving at some angle with respect to the other charges then the net force on the charge decreases with the increase in angle. Only the vertical component contributes to the force on the charge.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Two charges are placed at a certain distance apart class 12 physics CBSE
Difference Between Plant Cell and Animal Cell
What organs are located on the left side of your body class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The planet nearest to earth is A Mercury B Venus C class 6 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is BLO What is the full form of BLO class 8 social science CBSE