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Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be:
- a)5F
- b)F
- c)F/2
- d)F/5
Correct answer is option 'A'. Can you explain this answer?
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Two point charges placed in a medium of dielectric constant 5 are at a...
Q1 _______Q2
Force in the charges in the air is
= K F
= 5 F
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Two point charges placed in a medium of dielectric constant 5 are at a...
We need more information to calculate the electrostatic force between the two charges. We need to know the magnitude and sign of the charges. Once we have that information, we can use Coulomb's law:
F = k * q1 * q2 / r^2
where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
To take into account the medium of dielectric constant 5, we can replace k with the effective Coulomb's constant:
keff = k / ε
where ε is the dielectric constant of the medium.
So the electrostatic force between the two charges in the medium of dielectric constant 5 is:
F = keff * q1 * q2 / r^2
= k / ε * q1 * q2 / r^2
F = k * q1 * q2 / r^2
where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
To take into account the medium of dielectric constant 5, we can replace k with the effective Coulomb's constant:
keff = k / ε
where ε is the dielectric constant of the medium.
So the electrostatic force between the two charges in the medium of dielectric constant 5 is:
F = keff * q1 * q2 / r^2
= k / ε * q1 * q2 / r^2
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Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be:a)5Fb)Fc)F/2d)F/5Correct answer is option 'A'. Can you explain this answer?
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