The equilibrium position for a third charge can be found by considering the forces acting on it. The two fixed charges will exert a repulsive force on the third charge.
Let's assume the third charge is q and its position is x.
The force due to charge A on q can be calculated using Coulomb's Law:
Fa = (k * |q| * |A|) / x^2
where k is the Coulomb's constant (9 × 10^9 N m^2/C^2).
The force due to charge B on q can also be calculated using Coulomb's Law:
Fb = (k * |q| * |B|) / (0.25 - x)^2
Since q is in equilibrium, the forces Fa and Fb must be equal.
Therefore, we can set up an equation:
Fa = Fb
(k * |q| * |A|) / x^2 = (k * |q| * |B|) / (0.25 - x)^2
|A| / x^2 = |B| / (0.25 - x)^2
Now, we can substitute the given values of |A| and |B| into the equation:
2 / x^2 = 18 / (0.25 - x)^2
Now, we can simplify the equation:
(0.25 - x)^2 = (18 * x^2) / 2
(0.25 - x)^2 = 9 * x^2
Taking the square root of both sides:
0.25 - x = ±3 * x
Simplifying further:
0.25 = 4 * x
x = 0.25 / 4
x = 0.0625 m
Therefore, the third charge should be placed at a distance of 0.0625 m (or 6.25 cm) from charge A.
Two charges, A=+2nc & B=+18nc, are fixed 25 cm apart, as shown below. Where should a third charge be placed such that it would be in equilibrium?
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