two parallel plates are separated by a dielectric of thickness 2mm and a dielectric of constant 6.if the area of each plate is 40cm2 and p.d between them is 1.5kv,find the capacitance,the energy stored by the capacitor and the ratio of the capacitance if the dielectric is replaced by free space
To find the capacitance of the parallel plates, we can use the formula:
C = (ε₀ * εᵣ * A) / d
Where:
C is the capacitance,
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m),
εᵣ is the relative permittivity (dielectric constant) of the material,
A is the area of each plate, and
d is the separation between the plates.
Let's plug in the given values:
Relative permittivity, εᵣ = 6
Area, A = 40 cm² = (40/10000) m²
Separation, d = 2 mm = 0.002 m
Calculating the capacitance:
C = (8.85 x 10^-12 F/m) * 6 * (40/10000) m² / 0.002 m
C ≈ 0.212 F
The capacitance of the parallel plates is approximately 0.212 Farads.
To find the energy stored by the capacitor, we can use the formula:
E = (1/2) * C * V²
Where:
E is the energy stored,
C is the capacitance, and
V is the potential difference (voltage) between the plates.
Given:
Capacitance, C ≈ 0.212 F
Potential difference, V = 1.5 kV = 1500 V
Calculating the energy stored:
E = (1/2) * 0.212 F * (1500 V)²
E ≈ 190.5 J
The energy stored by the capacitor is approximately 190.5 Joules.
Now, let's calculate the ratio of the capacitance when the dielectric is replaced by free space.
C₁ is the capacitance with the dielectric, and C₂ is the capacitance with free space.
The ratio of the capacitance is given by:
Ratio = C₁ / C₂
When the dielectric is replaced by free space, the relative permittivity (εᵣ) is equal to 1.
Plugging in the values:
C₁ ≈ 0.212 F
C₂ = (ε₀ * ε₂ * A) / d
where ε₂ = 1
Ratio = C₁ / C₂ = 0.212 F / [(8.85 x 10^-12 F/m) * 1 * (40/10000) m² / 0.002 m]
Calculating the ratio:
Ratio ≈ 1 / 0.0446 ≈ 22.4
So, the ratio of the capacitance when the dielectric is replaced by free space is approximately 22.4.
To find the capacitance, you can use the formula:
C = (ε₀ × εᵣ × A) / d,
where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), εᵣ is the relative permittivity of the dielectric (given as 6), A is the area of each plate (40 cm^2 = 0.004 m^2), and d is the separation between the plates (2 mm = 0.002 m).
Plugging in these values, the formula becomes:
C = (8.85 x 10^-12 F/m × 6 × 0.004 m^2) / 0.002 m
= 0.0531 F (or 53.1 μF)
So, the capacitance of the capacitor is 53.1 microfarads.
To find the energy stored by the capacitor, you can use the formula:
E = (½) × C × V^2,
where E is the energy stored, C is the capacitance (0.0531 F), and V is the potential difference (1.5 kV = 1500 V).
Plugging in these values, the formula becomes:
E = (½) × 0.0531 F × (1500 V)^2
= 119.475 J (or 119.475 J)
So, the energy stored by the capacitor is 119.475 Joules.
Now, if the dielectric is replaced by free space, the relative permittivity (εᵣ) becomes 1. The formula for capacitance remains the same:
C = (ε₀ × εᵣ × A) / d,
where ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), εᵣ is now 1, A is the area of each plate (40 cm^2 = 0.004 m^2), and d is the separation between the plates (2 mm = 0.002 m).
Plugging in these values, the formula becomes:
C = (8.85 x 10^-12 F/m × 1 × 0.004 m^2) / 0.002 m
= 0.0177 F (or 17.7 μF)
So, the capacitance is 17.7 microfarads. The ratio of the capacitance when the dielectric is replaced by free space to the original capacitance is:
Ratio = Capacitance with free space / Original capacitance
= 17.7 μF / 53.1 μF
= 1/3
Therefore, the ratio of the capacitance when the dielectric is replaced by free space to the original capacitance is 1/3.