A parallel plate capacitor with plate area 40 cm2 and plate separation...
The electric field due to the polarization charges is
The electric field inside the dielectric slab is
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A parallel plate capacitor with plate area 40 cm2 and plate separation...
Given:
- Plate area (A) = 40 cm²
- Plate separation (d) = 5 mm = 0.5 cm
- Charge (Q) = 2 * 10^(-12) C
- Dielectric constant (k) = 5
- Dielectric thickness (t) = 3 mm = 0.3 cm
To find:
- Capacitance (C)
Formula:
The capacitance of a parallel plate capacitor with a dielectric is given by the formula:
C = (k * ε₀ * A) / d
where ε₀ is the permittivity of free space (ε₀ ≈ 8.854 * 10^(-12) F/m)
Solution:
Given that the plate area is 40 cm², we can convert it to m²:
A = 40 cm² = 40 * 10^(-4) m²
Similarly, the plate separation is 0.5 cm, which can be converted to meters:
d = 0.5 cm = 0.5 * 10^(-2) m
Now, let's calculate the capacitance without the dielectric:
C₁ = (ε₀ * A) / d
The permittivity of free space is approximately ε₀ ≈ 8.854 * 10^(-12) F/m, so substituting the values:
C₁ = (8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₁ = (8.854 * 40) / 0.5
C₁ = 708.32 F
Next, we need to calculate the capacitance with the dielectric:
C₂ = (k * ε₀ * A) / d
Substituting the values:
C₂ = (5 * 8.854 * 10^(-12) F/m * 40 * 10^(-4) m²) / (0.5 * 10^(-2) m)
C₂ = (5 * 8.854 * 40) / 0.5
C₂ = 1770.8 F
The capacitance of the capacitor with the dielectric is given by:
C = C₂ - C₁
C = 1770.8 F - 708.32 F
C ≈ 1062.48 F
However, the question asks for the capacitance in units of '11.4'. Therefore, the given answer seems to be incorrect. Please double-check the answer.