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Capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness, t = 2/3 d is introduced in between the plates. 'd' is the separation between the plates. The dielectric constant of the dielectric slab is
- a)14/11
- b)11/14
- c)7/11
- d)11/7
Correct answer is option 'A'. Can you explain this answer?
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Capacitance of a capacitor becomes 7/6 times its original value if a d...
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Given:
- Capacitance of the capacitor becomes 7/6 times its original value when a dielectric slab of thickness t = 2/3 d is introduced between the plates.
- d is the separation between the plates.
To find:
The dielectric constant of the dielectric slab.
Let's solve this step by step.
Step 1: Understanding the concept
The capacitance of a parallel plate capacitor is given by the formula:
C = ε₀A/d
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.
When a dielectric material is introduced between the plates, the capacitance increases by a factor of κ, which is the dielectric constant of the material. The new capacitance is given by:
C' = κε₀A/d
Step 2: Calculating the original capacitance
Let's assume the original capacitance is C₀.
C₀ = ε₀A/d
Step 3: Calculating the new capacitance
The capacitance becomes 7/6 times its original value, so the new capacitance is:
C' = (7/6)C₀
C' = (7/6)ε₀A/d
Step 4: Introducing the dielectric slab
When the dielectric slab is introduced, the thickness t = 2/3 d.
Therefore, the new separation between the plates is:
d' = d - t
Substituting the values in the formula for C':
C' = (7/6)ε₀A/d'
C' = (7/6)ε₀A/(d - t)
Step 5: Calculating the dielectric constant
We need to find the dielectric constant κ.
C' = κε₀A/d'
(7/6)C₀ = κε₀A/(d - t)
C₀ = κε₀A/d
Dividing both equations:
(7/6)C₀ / C₀ = κε₀A/(d - t) / (κε₀A/d)
7/6 = (d - t)/d
7d/6 = d - t
7d = 6d - 6t
6t = d
Substituting t = 2/3 d:
6(2/3 d) = d
4d = d
d = 4
Therefore, the separation between the plates is 4.
Step 6: Calculating the dielectric constant (contd.)
Substituting d = 4 in the equation:
6t = d
6t = 4
t = 4/6
t = 2/3
Therefore, the thickness of the dielectric slab is 2/3.
Step 7: Calculating the dielectric constant (contd.)
Substituting t = 2/3 and d = 4 in the equation:
7d/6 = d - t
7(4)/6 = 4 - 2/3
28/6 = 12/3 - 2/3
28/6 = 10/3
Cross-multiplying:
3(28) = 6(10)
84 = 60
This is a contradiction, so our assumption is incorrect.
Therefore, the dielectric constant of the
- Capacitance of the capacitor becomes 7/6 times its original value when a dielectric slab of thickness t = 2/3 d is introduced between the plates.
- d is the separation between the plates.
To find:
The dielectric constant of the dielectric slab.
Let's solve this step by step.
Step 1: Understanding the concept
The capacitance of a parallel plate capacitor is given by the formula:
C = ε₀A/d
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.
When a dielectric material is introduced between the plates, the capacitance increases by a factor of κ, which is the dielectric constant of the material. The new capacitance is given by:
C' = κε₀A/d
Step 2: Calculating the original capacitance
Let's assume the original capacitance is C₀.
C₀ = ε₀A/d
Step 3: Calculating the new capacitance
The capacitance becomes 7/6 times its original value, so the new capacitance is:
C' = (7/6)C₀
C' = (7/6)ε₀A/d
Step 4: Introducing the dielectric slab
When the dielectric slab is introduced, the thickness t = 2/3 d.
Therefore, the new separation between the plates is:
d' = d - t
Substituting the values in the formula for C':
C' = (7/6)ε₀A/d'
C' = (7/6)ε₀A/(d - t)
Step 5: Calculating the dielectric constant
We need to find the dielectric constant κ.
C' = κε₀A/d'
(7/6)C₀ = κε₀A/(d - t)
C₀ = κε₀A/d
Dividing both equations:
(7/6)C₀ / C₀ = κε₀A/(d - t) / (κε₀A/d)
7/6 = (d - t)/d
7d/6 = d - t
7d = 6d - 6t
6t = d
Substituting t = 2/3 d:
6(2/3 d) = d
4d = d
d = 4
Therefore, the separation between the plates is 4.
Step 6: Calculating the dielectric constant (contd.)
Substituting d = 4 in the equation:
6t = d
6t = 4
t = 4/6
t = 2/3
Therefore, the thickness of the dielectric slab is 2/3.
Step 7: Calculating the dielectric constant (contd.)
Substituting t = 2/3 and d = 4 in the equation:
7d/6 = d - t
7(4)/6 = 4 - 2/3
28/6 = 12/3 - 2/3
28/6 = 10/3
Cross-multiplying:
3(28) = 6(10)
84 = 60
This is a contradiction, so our assumption is incorrect.
Therefore, the dielectric constant of the
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Capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness, t = 2/3 d is introduced in between the plates. 'd' is the separation between the plates. The dielectric constant of the dielectric slab isa)14/11b)11/14c)7/11d)11/7Correct answer is option 'A'. Can you explain this answer?
Question Description
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