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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?
Options
(a) x/2
(b) x/3
(c) x/6
(d) 2x/3
Correct Answer:
x/3
Explanation:
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
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Topics: Electrostatics
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Subject: Physics
(2479)
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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 .