Question

two electron and a positive charge 'q'are held along straight line.at what position and for what value of 'q'will the system be in equilibrium?check whether it is stable ,unstable or neutral equilibrium

Answer 1

Answer:

Explanation:

See attachment

Answer 2

Answer:

The value of q is $2\times10^{-20}\ C$. This system is the stable.

Explanation:

Given that,

Two electron and a positive charge 'q'are held along straight line.

Let us consider a charge q be at the distances r₁ and r₂ from the two charges A and B respectively.

For equilibrium of q,

The forces on it exerted by A and B. The forces are equal and opposite.

The force on q by A

$F_{qA}=\dfrac{q}{4\pi\epsilon_{0}r_{1}^2}$....(I)

The force on q by B

$F_{qB}=\dfrac{q}{4\pi\epsilon_{0}r_{2}^2}$....(II)

Now, equate the equation (I) and (II)

$|F_{qA}|=|F_{qB}|$

$\dfrac{q}{4\pi\epsilon_{0}r_{1}^2}=\dfrac{q}{4\pi\epsilon_{0}r_{2}^2}$

$\dfrac{q}{r_{1}^2}=\dfrac{q}{r_{2}^2}$

let the distance is r₁=r₂=r.

Therefore, The charge q should be equidistant from A and B.

For the equilibrium system,

The charges A and B must also be in equilibrium.

The force on A by q is

$F_{1}=\dfrac{eq}{4\pi\epsilon_{0}r^2}$

The force on B by A is

$F_{2}=\dfrac{e^2}{4\pi\epsilon_{0}(2r)^2}$

The force on B by q is

$F_{3}=\dfrac{eq}{4\pi\epsilon_{0}r^2}$

For the equilibrium system

$F_{1} +F_{2}+F_{3}=0$

$\dfrac{eq}{r^2}+\dfrac{e^2}{2r^2}+\dfrac{eq}{r^2}=0$

$eq+\dfrac{e^2}{4}+eq=0$

$q = \dfrac{e}{8}$....(III)

Put the value of electron in equation (III)

$q=\dfrac{1.6\times10^{-19}}{8}$

$q = 2\times10^{-20}\ C$

Hence, The value of q is $2\times10^{-20}\ C$. This system is the stable.