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A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.
Correct answer is '5'. Can you explain this answer?
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A parallel plate capacitor is maintained at a certain potential differ...
If d is the separation between the plates (each of area A0) of a parallel plate condenser in air, then its capacitance in air, then its capacitance
If a slab of thickness t is introduced between the plates with new separation d’, then its new capacitance,
As Q = CV, the charge on the capacitor is same in both cases therefore to maintain the same potential difference the capacitance C and C’ must be same i.e, from (1) and (2)
Most Upvoted Answer
A parallel plate capacitor is maintained at a certain potential differ...
Given:
- A parallel plate capacitor maintained at a certain potential difference.
- Introduction of a 3 mm thick slab between the plates.
- In order to maintain the same potential difference, the distance between the plates is increased by 2.4 m.
To find:
The dielectric constant of the slab.
Explanation:
When a dielectric material is introduced between the plates of a capacitor, it affects the capacitance of the capacitor. The capacitance of a parallel plate capacitor with a dielectric material is given by the equation:
C = (ε₀ * εᵣ * A) / d,
where:
C is the capacitance,
ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m),
εᵣ is the relative permittivity or dielectric constant of the slab,
A is the area of the plates, and
d is the distance between the plates.
Step 1: Find the initial capacitance without the slab.
Since the potential difference across the capacitor remains the same, the initial capacitance (C₀) can be represented as:
C₀ = (ε₀ * A) / d₀,
where d₀ is the initial distance between the plates.
Step 2: Find the final capacitance with the slab.
After introducing the slab, the distance between the plates increases to (d₀ + 2.4 m). Therefore, the final capacitance (C) can be represented as:
C = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).
Step 3: Equate the initial and final capacitance.
Since the potential difference remains the same, we can equate C₀ and C:
(ε₀ * A) / d₀ = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).
Step 4: Solve the equation for εᵣ.
By cross-multiplying and rearranging the equation, we get:
(ε₀ * A) * (d₀ + 2.4 m) = (ε₀ * εᵣ * A) * d₀.
Simplifying further:
d₀ + 2.4 m = εᵣ * d₀.
Dividing both sides by d₀:
1 + (2.4 m / d₀) = εᵣ.
Step 5: Substitute the given values and calculate εᵣ.
The given thickness of the slab is 3 mm, which is equivalent to 0.003 m. The increase in distance between the plates is 2.4 m. Substituting these values into the equation:
1 + (2.4 m / d₀) = εᵣ,
1 + (2.4 m / d₀) = 5.
Therefore, the dielectric constant of the slab is 5.
- A parallel plate capacitor maintained at a certain potential difference.
- Introduction of a 3 mm thick slab between the plates.
- In order to maintain the same potential difference, the distance between the plates is increased by 2.4 m.
To find:
The dielectric constant of the slab.
Explanation:
When a dielectric material is introduced between the plates of a capacitor, it affects the capacitance of the capacitor. The capacitance of a parallel plate capacitor with a dielectric material is given by the equation:
C = (ε₀ * εᵣ * A) / d,
where:
C is the capacitance,
ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m),
εᵣ is the relative permittivity or dielectric constant of the slab,
A is the area of the plates, and
d is the distance between the plates.
Step 1: Find the initial capacitance without the slab.
Since the potential difference across the capacitor remains the same, the initial capacitance (C₀) can be represented as:
C₀ = (ε₀ * A) / d₀,
where d₀ is the initial distance between the plates.
Step 2: Find the final capacitance with the slab.
After introducing the slab, the distance between the plates increases to (d₀ + 2.4 m). Therefore, the final capacitance (C) can be represented as:
C = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).
Step 3: Equate the initial and final capacitance.
Since the potential difference remains the same, we can equate C₀ and C:
(ε₀ * A) / d₀ = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).
Step 4: Solve the equation for εᵣ.
By cross-multiplying and rearranging the equation, we get:
(ε₀ * A) * (d₀ + 2.4 m) = (ε₀ * εᵣ * A) * d₀.
Simplifying further:
d₀ + 2.4 m = εᵣ * d₀.
Dividing both sides by d₀:
1 + (2.4 m / d₀) = εᵣ.
Step 5: Substitute the given values and calculate εᵣ.
The given thickness of the slab is 3 mm, which is equivalent to 0.003 m. The increase in distance between the plates is 2.4 m. Substituting these values into the equation:
1 + (2.4 m / d₀) = εᵣ,
1 + (2.4 m / d₀) = 5.
Therefore, the dielectric constant of the slab is 5.
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Community Answer
A parallel plate capacitor is maintained at a certain potential differ...
As here with introduction of slab or with change in separation, potential difference between the plates is changed, the charged capacitor is isolated, i.e., q = q'
So, C0V0 = CV [as q = CV]
or, C0 = C [as V = V0 given]
Now as with introduction of dielectric slab the capacity increases from
by increasing d to (d + 0.24 cm) the capacity C1, becomes
So, C0V0 = CV [as q = CV]
or, C0 = C [as V = V0 given]
Now as with introduction of dielectric slab the capacity increases from
by increasing d to (d + 0.24 cm) the capacity C1, becomes
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A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer?
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A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer?.
A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.Correct answer is '5'. Can you explain this answer?.
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