# Two charges 4e and e are at a distance 'x' apart. At what distance a charge 'q' must be placed from charge e so that it is in equilibrium?

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e ......q.......4e

_r___

______x___

Force on q due to e

= k qe / r^2

Force on q due to 4e

= k q4e / ( x - r )^2

For it to be in equibrium :

The forces must be equal and opposite .

kqe/r^2 = k4eq/(x-r)^2

=> 1/r^2 = 4/(x-r)^2

=> x^2 + r^2 -2xr = 4r^2

=> 3r^2 + 2xr -x^2 = 0

=> 3r^2 + 3xr - xr - x^2 = 0

=> 3r ( r + x ) -x ( r + x ) = 0

=> ( 3r-x ) ( r + x ) = 0

=> r = -x or r = x/3

Since distance is non-negative

r = x/3 should be the answer .

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Let the distance of q from e is a units

**so, to be in equilibrium, the net force must be zero.**

**Force experienced on q is due to e and 4e**

Electrostatic force = (k Q1 ×Q2)/r²

A/q,

(k q×4e)/(x- a)² = (k q ×e)/a²

⇒4a² = (x -a)²

⇒x² -2ax - 3a²=0

⇒x² - 3ax + ax -3a² =0

⇒x(x- 3a) +a(x -3a)=0

⇒(x+a)(x-3a)=0

⇒x = -a or x= 3a

or, a= -x or a= x/3

**a can't be negative as q is placed in between e and 4e**

**so, a = x/3 units from charge e**

Electrostatic force = (k Q1 ×Q2)/r²

A/q,

(k q×4e)/(x- a)² = (k q ×e)/a²

⇒4a² = (x -a)²

⇒x² -2ax - 3a²=0

⇒x² - 3ax + ax -3a² =0

⇒x(x- 3a) +a(x -3a)=0

⇒(x+a)(x-3a)=0

⇒x = -a or x= 3a

or, a= -x or a= x/3