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Two identical metal spheres charged with +12µF and - 8µF are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is :
- a)12 : 1
- b)8 : 1
- c)24 : 1
- d)4 : 1
Correct answer is option 'C'. Can you explain this answer?
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Two identical metal spheres charged with +12µF and - 8µF are...
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Two identical metal spheres charged with +12µF and - 8µF are...
Given: Two identical metal spheres charged with 12F and -8F
To find: Ratio of the magnitudes of electrostatic forces between them before and after contact
Solution:
1. Coulomb's Law: The force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
2. Let the distance between the two spheres be 'd'.
3. Magnitude of electrostatic force between the spheres before contact:
From Coulomb's Law,
F1 = (1/4πε) * ((12F) * (-8F))/d^2
where ε is the permittivity of air.
F1 = -3 * 10^9/d^2
Note: The negative sign indicates that the force is repulsive.
4. After the spheres are brought into contact, they will share the charges uniformly.
5. Charge on each sphere after contact = (12F - 8F)/2 = 2F
6. Magnitude of electrostatic force between the spheres after contact:
From Coulomb's Law,
F2 = (1/4πε) * ((2F) * (2F))/d^2
where ε is the permittivity of air.
F2 = 5 * 10^9/d^2
7. Ratio of the magnitudes of electrostatic forces between the spheres before and after contact:
F1/F2 = (-3 * 10^9/d^2) / (5 * 10^9/d^2) = -3/5
Taking the absolute value, the ratio of the magnitudes of electrostatic forces between the spheres before and after contact is 3:5.
8. Simplifying the ratio:
3/5 = 9/15 = 12/20 = 24/40
Therefore, the correct option is (C) 24:1.
To find: Ratio of the magnitudes of electrostatic forces between them before and after contact
Solution:
1. Coulomb's Law: The force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
2. Let the distance between the two spheres be 'd'.
3. Magnitude of electrostatic force between the spheres before contact:
From Coulomb's Law,
F1 = (1/4πε) * ((12F) * (-8F))/d^2
where ε is the permittivity of air.
F1 = -3 * 10^9/d^2
Note: The negative sign indicates that the force is repulsive.
4. After the spheres are brought into contact, they will share the charges uniformly.
5. Charge on each sphere after contact = (12F - 8F)/2 = 2F
6. Magnitude of electrostatic force between the spheres after contact:
From Coulomb's Law,
F2 = (1/4πε) * ((2F) * (2F))/d^2
where ε is the permittivity of air.
F2 = 5 * 10^9/d^2
7. Ratio of the magnitudes of electrostatic forces between the spheres before and after contact:
F1/F2 = (-3 * 10^9/d^2) / (5 * 10^9/d^2) = -3/5
Taking the absolute value, the ratio of the magnitudes of electrostatic forces between the spheres before and after contact is 3:5.
8. Simplifying the ratio:
3/5 = 9/15 = 12/20 = 24/40
Therefore, the correct option is (C) 24:1.
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Two identical metal spheres charged with +12µF and - 8µF are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is :a)12 : 1b)8 : 1c)24 : 1d)4 : 1Correct answer is option 'C'. Can you explain this answer?
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