Two charges 4 µC and -16 µC are fixed and separated by a distance of 0.6 m. At what distance from charge 4 µC should a third charge 6 µC be placed so that it will not experience an... Two charges 4 µC and -16 µC are fixed and separated by a distance of 0.6 m. At what distance from charge 4 µC should a third charge 6 µC be placed so that it will not experience any force?
Understand the Problem
The question is asking to determine the distance from a positive charge where a third charge will not experience any force due to two other fixed charges. This involves concepts from electrostatics and requires understanding the forces between charged particles.
Answer
The third charge should be placed at a distance of $0.4 \, m$ from the positive charge.
Answer for screen readers
The valid distance for charge ( q_3 ) is ( x \approx 0.4 , m ).
Steps to Solve
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Identify the charges and their positions
Let the two charges be ( q_1 = 4 , \mu C ) and ( q_2 = -16 , \mu C ) situated 0.6 m apart. Assume charge ( q_3 = 6 , \mu C ) is to be placed between them.
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Set up the position of charge ( q_3 )
Let the distance from ( q_1 ) to ( q_3 ) be ( x ). Hence, the distance from ( q_2 ) to ( q_3 ) would be ( 0.6 - x ).
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Calculate the forces on charge ( q_3 )
The force due to ( q_1 ) acting on ( q_3 ) is given by Coulomb's law:
$$ F_1 = k \frac{|q_1 \cdot q_3|}{x^2} $$
The force due to ( q_2 ) should be equal and opposite for the charge ( q_3 ) to experience no net force:
$$ F_2 = k \frac{|q_2 \cdot q_3|}{(0.6 - x)^2} $$
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Set the forces equal to each other
For charge ( q_3 ) to experience no net force, set ( F_1 = F_2 ):
$$ k \frac{|4 \cdot 6|}{x^2} = k \frac{|-16 \cdot 6|}{(0.6 - x)^2} $$
Since ( k ) can be canceled out, we get:
$$ \frac{24}{x^2} = \frac{96}{(0.6 - x)^2} $$
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Cross-multiply and simplify
Cross-multiplying gives:
$$ 24(0.6 - x)^2 = 96x^2 $$
Expanding this gives:
$$ 24(0.36 - 1.2x + x^2) = 96x^2 $$
Therefore:
$$ 8.64 - 28.8x + 24x^2 = 96x^2 $$
Rearranging results in:
$$ 72x^2 + 28.8x - 8.64 = 0 $$
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Solve the quadratic equation
The equation can be solved using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where:
- ( a = 72 )
- ( b = 28.8 )
- ( c = -8.64 )
Calculate the discriminant:
$$ D = b^2 - 4ac = (28.8)^2 - 4(72)(-8.64) $$
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Calculate the roots
After calculating ( D ), find:
$$ x = \frac{-28.8 \pm \sqrt{D}}{2 \cdot 72} $$
Solve for the two potential distances for ( x ).
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Determine which root is valid
Only keep roots that satisfy ( 0 < x < 0.6 ).
The valid distance for charge ( q_3 ) is ( x \approx 0.4 , m ).
More Information
Forces between charges depend on their magnitude and distance apart. The positions are critical for achieving an equilibrium where the net force is zero, which is common in electrostatics problems.
Tips
- Misunderstanding the direction of forces; always consider attractive vs. repulsive forces.
- Neglecting to check that distances are valid (between the two fixed charges).
- Not simplifying the quadratic equation properly, leading to incorrect roots.
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