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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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Given:
Original capacitance of the capacitor = C
Capacitance after introducing the dielectric slab = (7/6)C
Thickness of the dielectric slab = t = 2/3d

To find: Dielectric constant of the dielectric slab (let it be K)

Formula:
The formula for the capacitance of a capacitor with a dielectric slab is given by:

C' = (K * ε₀ * A) / d

Where,
C' = Capacitance with the dielectric slab
K = Dielectric constant
ε₀ = Permittivity of free space
A = Area of the capacitor plates
d = Separation between the plates

1. Calculate the new capacitance:
C' = (7/6)C

2. Calculate the original capacitance:
C = C' * (6/7)

3. Substitute the value of C' in the formula:
(7/6)C = (K * ε₀ * A) / d

4. Substitute the value of C from step 2:
(7/6) * (6/7)C = (K * ε₀ * A) / d

5. Simplify the equation:
C = (K * ε₀ * A) / d

6. Substitute the given value of t = 2/3d:
C = (K * ε₀ * A) / (2/3d)

7. Simplify the equation further:
C = (3K * ε₀ * A) / 2d

8. Compare the equation with the original capacitance formula (without the dielectric slab):
C = (ε₀ * A) / d

9. Equate the two equations from step 7 and step 8:
(3K * ε₀ * A) / 2d = (ε₀ * A) / d

10. Cancel out common terms:
3K / 2 = 1

11. Solve for K:
K = 2/3 * 1

K = 2/3

Therefore, the dielectric constant of the dielectric slab is 2/3, which is equal to 14/21 in fraction form.
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