Question

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?

Answer 1

Let q be r distant from e .

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 .

Answer 2

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