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Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r/2, then the force between the charges will become
- a)F
- b)3/5 F
- c)4/9 F
- d)F/4
- e)F/2
Correct answer is option 'C'. Can you explain this answer?
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Concept:
The force between two charges in vacuum is given by Coulomb's law:
F = k * (q1 * q2) / r^2
where F is the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
When a dielectric is inserted between the charges, the force between them is modified due to the presence of the dielectric. The force is given by:
F' = k' * (q1 * q2) / r^2
where k' is the modified electrostatic constant, which is equal to k multiplied by the relative permittivity of the dielectric.
Given:
The force between the charges in vacuum is F.
The dielectric constant of the slab is 4.
The thickness of the slab is r/2.
Calculation:
Step 1: Find the modified electrostatic constant (k').
k' = k * relative permittivity
The relative permittivity of the dielectric is given by:
relative permittivity = dielectric constant
So, k' = k * 4
Step 2: Find the force between the charges with the dielectric (F').
F' = k' * (q1 * q2) / r^2
Since the charges are identical, q1 = q2 = q, so we can write:
F' = k' * (q * q) / r^2
Step 3: Compare the force F' with the original force F.
F' / F = (k' * (q * q) / r^2) / (k * (q * q) / r^2)
F' / F = (k' / k) * (q * q / q * q) * (r^2 / r^2)
F' / F = (k' / k) * 1
F' / F = k' / k
Substituting the values of k' and k, we get:
F' / F = (4 * k) / k
F' / F = 4
Therefore, the force between the charges with the dielectric is 4 times the original force.
Answer:
The force between the charges will become 4/9 F. (Option C)
The force between two charges in vacuum is given by Coulomb's law:
F = k * (q1 * q2) / r^2
where F is the force, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
When a dielectric is inserted between the charges, the force between them is modified due to the presence of the dielectric. The force is given by:
F' = k' * (q1 * q2) / r^2
where k' is the modified electrostatic constant, which is equal to k multiplied by the relative permittivity of the dielectric.
Given:
The force between the charges in vacuum is F.
The dielectric constant of the slab is 4.
The thickness of the slab is r/2.
Calculation:
Step 1: Find the modified electrostatic constant (k').
k' = k * relative permittivity
The relative permittivity of the dielectric is given by:
relative permittivity = dielectric constant
So, k' = k * 4
Step 2: Find the force between the charges with the dielectric (F').
F' = k' * (q1 * q2) / r^2
Since the charges are identical, q1 = q2 = q, so we can write:
F' = k' * (q * q) / r^2
Step 3: Compare the force F' with the original force F.
F' / F = (k' * (q * q) / r^2) / (k * (q * q) / r^2)
F' / F = (k' / k) * (q * q / q * q) * (r^2 / r^2)
F' / F = (k' / k) * 1
F' / F = k' / k
Substituting the values of k' and k, we get:
F' / F = (4 * k) / k
F' / F = 4
Therefore, the force between the charges with the dielectric is 4 times the original force.
Answer:
The force between the charges will become 4/9 F. (Option C)
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