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Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.
- a)1.44
- b)2.44
- c)1.404
- d)2.404
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Find the force between two charges when they are brought in contact an...
Order to find the force between two charges when they are brought in contact and then separated, we need to consider the concept of charge conservation. When two charges are brought in contact, they will share their charges in such a way that the total charge remains conserved.
Let's assume the initial charges of the two objects are q1 = 2nC and q2 = -1nC. When they are brought in contact, the charges will redistribute themselves to reach a state of equilibrium.
Since the total charge must remain conserved, the final charges can be calculated using the equation:
q1 + q2 = q1_final + q2_final
Substituting the initial charges:
2nC + (-1nC) = q1_final + q2_final
1nC = q1_final + q2_final
Since the charges are redistributed equally when two objects are brought in contact, q1_final = q2_final = 0.5nC.
Now, the force between two charges can be calculated using Coulomb's Law:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the electrostatic constant (9 × 10^9 N m²/C²), q1 and q2 are the charges, and r is the distance between the charges.
Substituting the values:
F = (9 × 10^9 N m²/C² * |2nC * (-1nC)|) / (0.04m)^2
F = (9 × 10^9 N m²/C² * 2nC * 1nC) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1nC²) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1 × 10^-9C²) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1 × 10^-9) / 0.0016 N
F = 11.25 × 10^9 N
Therefore, the force between the two charges when they are brought in contact and separated by 4cm apart is 11.25 × 10^9 N.
Let's assume the initial charges of the two objects are q1 = 2nC and q2 = -1nC. When they are brought in contact, the charges will redistribute themselves to reach a state of equilibrium.
Since the total charge must remain conserved, the final charges can be calculated using the equation:
q1 + q2 = q1_final + q2_final
Substituting the initial charges:
2nC + (-1nC) = q1_final + q2_final
1nC = q1_final + q2_final
Since the charges are redistributed equally when two objects are brought in contact, q1_final = q2_final = 0.5nC.
Now, the force between two charges can be calculated using Coulomb's Law:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the electrostatic constant (9 × 10^9 N m²/C²), q1 and q2 are the charges, and r is the distance between the charges.
Substituting the values:
F = (9 × 10^9 N m²/C² * |2nC * (-1nC)|) / (0.04m)^2
F = (9 × 10^9 N m²/C² * 2nC * 1nC) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1nC²) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1 × 10^-9C²) / 0.0016m²
F = (18 × 10^9 N m²/C² * 1 × 10^-9) / 0.0016 N
F = 11.25 × 10^9 N
Therefore, the force between the two charges when they are brought in contact and separated by 4cm apart is 11.25 × 10^9 N.
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Community Answer
Find the force between two charges when they are brought in contact an...
Before the charges are brought into contact, F = 11.234 μN.
After charges are brought into contact and then separated, charge on each sphere is, (q1 + q2)/2 = 0.5nC
On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN.
After charges are brought into contact and then separated, charge on each sphere is, (q1 + q2)/2 = 0.5nC
On calculating the force with q1 = q2 = 0.5nC, F = 1.404μN.
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Find the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN.a)1.44b)2.44c)1.404d)2.404Correct answer is option 'C'. Can you explain this answer?
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