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The plates of a parallel plate capacitor are charged upto 200 volts. A dielectric slab of thickness 4 mm is inserted between its plates. Then to maintain the same potential difference between the plates capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of dielectric slab is
- a)1
- b)4
- c)5
- d)6
Correct answer is option 'C'. Can you explain this answer?
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The plates of a parallel plate capacitor are charged upto 200 volts. A...
Given data:
Voltage (V) = 200 V
Thickness of dielectric slab (d) = 4 mm
Increase in distance between plates (Δd) = 3.2 mm
Formula:
The formula for calculating the capacitance of a parallel plate capacitor with a dielectric slab is:
C = (ε0 * εr * A) / d
Explanation:
1. Initial capacitance (C1):
Initially, when the dielectric slab is not inserted, the capacitance of the parallel plate capacitor is given by:
C1 = (ε0 * A) / d
Where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
2. Final capacitance with dielectric (C2):
When the dielectric slab is inserted, the capacitance becomes:
C2 = (ε0 * εr * A) / d
Where εr is the relative permittivity or dielectric constant of the material.
3. Change in capacitance:
The change in capacitance ΔC = C2 - C1 = (ε0 * A / d) * (εr - 1)
4. Change in distance:
Given that the distance between the plates is increased by 3.2 mm, the new distance becomes d' = d + Δd
5. Equating the potential difference:
To maintain the same potential difference between the plates, the charge on the plates remains the same. Therefore, the product of capacitance and potential difference remains constant:
C1 * V = C2 * V'
Where V is the initial potential difference and V' is the final potential difference.
6. Calculating the dielectric constant:
Substitute the values in the equation and calculate the dielectric constant:
C1 * V = C2 * V'
(ε0 * A / d) * V = (ε0 * εr * A / d') * V
εr = (d / d') = (4 / 7.2) = 5
Therefore, the dielectric constant of the dielectric slab is 5. Hence, the correct answer is option c) 5.
Voltage (V) = 200 V
Thickness of dielectric slab (d) = 4 mm
Increase in distance between plates (Δd) = 3.2 mm
Formula:
The formula for calculating the capacitance of a parallel plate capacitor with a dielectric slab is:
C = (ε0 * εr * A) / d
Explanation:
1. Initial capacitance (C1):
Initially, when the dielectric slab is not inserted, the capacitance of the parallel plate capacitor is given by:
C1 = (ε0 * A) / d
Where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
2. Final capacitance with dielectric (C2):
When the dielectric slab is inserted, the capacitance becomes:
C2 = (ε0 * εr * A) / d
Where εr is the relative permittivity or dielectric constant of the material.
3. Change in capacitance:
The change in capacitance ΔC = C2 - C1 = (ε0 * A / d) * (εr - 1)
4. Change in distance:
Given that the distance between the plates is increased by 3.2 mm, the new distance becomes d' = d + Δd
5. Equating the potential difference:
To maintain the same potential difference between the plates, the charge on the plates remains the same. Therefore, the product of capacitance and potential difference remains constant:
C1 * V = C2 * V'
Where V is the initial potential difference and V' is the final potential difference.
6. Calculating the dielectric constant:
Substitute the values in the equation and calculate the dielectric constant:
C1 * V = C2 * V'
(ε0 * A / d) * V = (ε0 * εr * A / d') * V
εr = (d / d') = (4 / 7.2) = 5
Therefore, the dielectric constant of the dielectric slab is 5. Hence, the correct answer is option c) 5.
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The plates of a parallel plate capacitor are charged upto 200 volts. A dielectric slab of thickness 4 mm is inserted between its plates. Then to maintain the same potential difference between the plates capacitor, the distance between the plates is increased by 3.2 mm. The dielectric constant of dielectric slab isa)1b)4c)5d)6Correct answer is option 'C'. Can you explain this answer?
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