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Two point charges in air at a distance of 20 cm. from each other interact with a certain force. At what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction
- a)8.94 × 10-2 m
- b)0.894 × 10-2 m
- c)89.4 × 10-2 m
- d)8.94 × 102 m
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Two point charges in air at a distance of 20 cm. from each other inter...
d=8.94 cm or
d=8.94x10-2
d=8.94x10-2
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Two point charges in air at a distance of 20 cm. from each other inter...
We can use Coulomb's law to relate the force of interaction between the charges to their separation distance and the permittivity of the medium:
F = (1/4πε) * q1 * q2 / r^2
where F is the force of interaction, ε is the permittivity of the medium, q1 and q2 are the magnitudes of the charges, and r is the separation distance.
We want to find the new separation distance r' in the oil such that the force of interaction is the same. Let's set up a proportion:
F / ε = F' / ε'
where F is the original force of interaction in air, ε is the permittivity of air (which is approximately 1), F' is the desired force of interaction in oil, and ε' is the permittivity of oil (which is given as 5).
We can rearrange this equation to solve for r':
F' = F * ε' / ε = (1/4π) * q1 * q2 * (ε' / ε) / r'^2
r'^2 = (1/4π) * q1 * q2 * (ε / ε') * (1 / F')
Plugging in the values given in the problem, we get:
r'^2 = (1/4π) * (2 μC)^2 * (ε / ε') * (1 / F) = (1/4π) * (2 × 10^-6 C)^2 * (1 / 5) * (1 / F) = 8.94 cm^2
Taking the square root of both sides, we get:
r' = 2.99 cm
Therefore, the two charges should be placed 2.99 cm apart in the oil to obtain the same force of interaction as in air.
F = (1/4πε) * q1 * q2 / r^2
where F is the force of interaction, ε is the permittivity of the medium, q1 and q2 are the magnitudes of the charges, and r is the separation distance.
We want to find the new separation distance r' in the oil such that the force of interaction is the same. Let's set up a proportion:
F / ε = F' / ε'
where F is the original force of interaction in air, ε is the permittivity of air (which is approximately 1), F' is the desired force of interaction in oil, and ε' is the permittivity of oil (which is given as 5).
We can rearrange this equation to solve for r':
F' = F * ε' / ε = (1/4π) * q1 * q2 * (ε' / ε) / r'^2
r'^2 = (1/4π) * q1 * q2 * (ε / ε') * (1 / F')
Plugging in the values given in the problem, we get:
r'^2 = (1/4π) * (2 μC)^2 * (ε / ε') * (1 / F) = (1/4π) * (2 × 10^-6 C)^2 * (1 / 5) * (1 / F) = 8.94 cm^2
Taking the square root of both sides, we get:
r' = 2.99 cm
Therefore, the two charges should be placed 2.99 cm apart in the oil to obtain the same force of interaction as in air.
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