Two charges 1 μC and 9 μC are placed at separation of 1 m. Where should a third charge be placed such that, it experience no net force due to these charges?
Answers
Given:
Magnitude of charge 1 (q₁) = 1 μC
Magnitude of charge 2 (q₂) = 9 μC
Distance of separation = 1 m
To find:
The position at which a third charge be placed such that it experience no force
Calculation :
Let the magnitude of third charge be q₃ and it is placed at a distance of ‘x’ m from charge 1. Then the distance of separation between charge 2 and charge 3 will be (1-x) m
Force between two charged particles is given by the formula
F = kq₁q₂/r²
Force between the charges 1 & 3 = kq₁q₃/x²
Force between the charges 2 & 3 = kq₂q₃ / (x-1)²
Resultant force at third charge = 0
=> kq₂q₃ / (x-1)² - kq₁q₃/x² = 0
=> kq₂q₃ / (x-1)² = kq₁q₃/x²
=> q₂/(x-1)² = q₁/x²
=> 9/(x-1)² = 1/x²
=> 9x² = x²-2x+1
=> 8x²+2x-1 = 0
=> 8x²+4x-2x-1 = 0
=> 4x(2x+1) - 1(2x+1) = 0
=> (4x-1)(2x+1) = 0
As x cannot be negative
=> 4x-1=0
=> x=1/4
The third charge should be placed at a distance of 0.25m from 1 μC charge
Answer:
3/4 m away from 9μC.
1/4 m away from 1 μC
Explanation:
Force = k q₁q₂/r², where q₁ and q₂ are charges and r is distance between them.
Here, let the third should be placed at x distance form 9μC charge. Hence, 1 - x distance from other one.
As these charges are +, that must be -ve , to attract them, so that the net force become 0. Let the be q.
⇒ k 9μ*q/x^2 = k 1μ*q/(1 - x)^2
⇒ 9/x^2 = 1/(1 - x)^2
⇒ 9(1 + x^2 - 2x) = x^2
⇒ 8x^2 - 18x + 9 = 0
⇒ 8x^2 - 6x - 12x + 9 = 0
⇒ (2x - 3)(4x - 3) = 0
⇒ x = 3/2 or 3/4 = x
There are two such places. Placed either at the distance of 3/2 m from 9μC or at 3/4 from 9μC. However if put it outside, it will be attracted by the near charge. And hence, 3/4 m is correct.