| ⇦ | 
| ⇨ |
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
Related Questions: - The moment of inertia of a thin uniform rod of mass M and length L about an axis
- Parsec is the unit of
- A battery of e.m.f. E and internal resistance r is connected to an external resistance
- Radiofrequency choke uses core of
- A cylindrical capacitors has charge Q and length L. If both the charge and length
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The moment of inertia of a thin uniform rod of mass M and length L about an axis
- Parsec is the unit of
- A battery of e.m.f. E and internal resistance r is connected to an external resistance
- Radiofrequency choke uses core of
- A cylindrical capacitors has charge Q and length L. If both the charge and length
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
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 .