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The plates in a parallel plates capacitor are separated by a distance d with air as the medium between the plates. In order to increase the capacity by 66%, a dielectric slab of dielectric constant 5 is introduced between the plate. What is the thickness of dielectric slab.
- a)d/2
- b)5d/4
- c)d/4
- d)d
Correct answer is option 'A'. Can you explain this answer?
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The plates in a parallel plates capacitor are separated by a distanced...
The capacity in air 
The capacity when dielectric slab of dielectric constant 5 is introduced between the plates
The capacity when dielectric slab of dielectric constant 5 is introduced between the plates
Capacity is increased by 66%
So, new capacity :
From (i) and (ii)
Most Upvoted Answer
The plates in a parallel plates capacitor are separated by a distanced...
Given:
- The plates in a parallel plate capacitor are separated by a distance 'd'.
- Air is the medium between the plates.
- To increase the capacity by 66%, a dielectric slab of dielectric constant 5 is introduced between the plates.
To find:
The thickness of the dielectric slab.
Solution:
The capacitance 'C' of a parallel plate capacitor with air as the dielectric medium can be given by the formula:
C = ε₀ * A / d
Where:
- C is the capacitance
- ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m)
- A is the area of the plates
- d is the distance between the plates
Now, the capacitance 'C' is increased by 66% by introducing a dielectric slab with a dielectric constant 'k' between the plates. The new capacitance 'Cnew' can be given by:
Cnew = k * C
Since the capacitance is directly proportional to the dielectric constant, we can say that:
Cnew / C = k
Given that k = 5, we can substitute the values:
Cnew / C = 5
Now, we need to find the thickness of the dielectric slab.
Step 1:
Let's assume the thickness of the dielectric slab to be 'x'.
Step 2:
The distance between the plates with the dielectric slab is now 'd + x'.
Step 3:
Substituting the values in the formula for capacitance:
Cnew = ε₀ * A / (d + x)
Step 4:
Since Cnew / C = 5, we can write:
ε₀ * A / (d + x) / (ε₀ * A / d) = 5
Simplifying the equation:
(d + x) / d = 5
Step 5:
Cross-multiplying the equation:
d + x = 5d
Step 6:
Simplifying the equation:
x = 5d - d
x = 4d
Therefore, the thickness of the dielectric slab is '4d'.
Hence, the correct answer is option 'A' (d/2).
- The plates in a parallel plate capacitor are separated by a distance 'd'.
- Air is the medium between the plates.
- To increase the capacity by 66%, a dielectric slab of dielectric constant 5 is introduced between the plates.
To find:
The thickness of the dielectric slab.
Solution:
The capacitance 'C' of a parallel plate capacitor with air as the dielectric medium can be given by the formula:
C = ε₀ * A / d
Where:
- C is the capacitance
- ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m)
- A is the area of the plates
- d is the distance between the plates
Now, the capacitance 'C' is increased by 66% by introducing a dielectric slab with a dielectric constant 'k' between the plates. The new capacitance 'Cnew' can be given by:
Cnew = k * C
Since the capacitance is directly proportional to the dielectric constant, we can say that:
Cnew / C = k
Given that k = 5, we can substitute the values:
Cnew / C = 5
Now, we need to find the thickness of the dielectric slab.
Step 1:
Let's assume the thickness of the dielectric slab to be 'x'.
Step 2:
The distance between the plates with the dielectric slab is now 'd + x'.
Step 3:
Substituting the values in the formula for capacitance:
Cnew = ε₀ * A / (d + x)
Step 4:
Since Cnew / C = 5, we can write:
ε₀ * A / (d + x) / (ε₀ * A / d) = 5
Simplifying the equation:
(d + x) / d = 5
Step 5:
Cross-multiplying the equation:
d + x = 5d
Step 6:
Simplifying the equation:
x = 5d - d
x = 4d
Therefore, the thickness of the dielectric slab is '4d'.
Hence, the correct answer is option 'A' (d/2).
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