If two charges +4e and +e are at a distance x apart, then at what distance charge q

If two charges +4e and +e are at a distance x apart, then at what distance charge q must be placed from +e, so that it is in equilibrium?

(a) x/2
(b) x/3
(c) x/6
(d) 2x/3

x/3

For equilibrium of q, |F₁| = |F₂|

(1 / 4πε₀).(qq₁ / x₁) = (1 / 4πε₀).(qq₂ / x₂)

x₂ = x / √[(q₁/q₂) + 1] = x / √[(4e/e) + 1] = x / 3

admin:

Pratham Parle says:

February 18, 2018 at 1:35 pm

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 .