Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k = 3.9). The empty capacitor is connected to a 17 -V battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?
To determine the potential difference across the plates of the capacitor filled with a dielectric, we can use the formula for the energy stored in a capacitor.
The energy stored in a capacitor is given by the equation:
E = (1/2) * C * V^2
Where:
E is the energy stored
C is the capacitance of the capacitor
V is the potential difference across the plates
The capacitance of a capacitor is given by the equation:
C = (k * ε₀ * A) / d
Where:
k is the dielectric constant
ε₀ is the permittivity of free space (8.85 x 10^-12 F/m)
A is the area of the plates
d is the distance between the plates
Since we have two identical capacitors, the area and distance between the plates are the same. Let's assume A and d are constant.
For the empty capacitor (C₁):
C₁ = ε₀ * A / d
For the capacitor filled with a dielectric (C₂):
C₂ = (k * ε₀ * A) / d
We want both capacitors to store the same amount of electrical energy. Therefore, we can equate the energies:
E₁ = E₂
(1/2) * C₁ * V₁^2 = (1/2) * C₂ * V₂^2
Plugging in the expressions for C₁ and C₂:
(1/2) * (ε₀ * A / d) * V₁^2 = (1/2) * ((k * ε₀ * A) / d) * V₂^2
The area and distance between the plates cancel out, giving us:
ε₀ * V₁^2 = k * ε₀ * V₂^2
Dividing both sides by ε₀:
V₁^2 = k * V₂^2
Taking the square root of both sides:
V₁ = √(k) * V₂
Substituting the given values:
V₁ = √(3.9) * V₂
Since the potential difference of the empty capacitor (V₁) is given as 17 V, we can solve for V₂:
17 V = √(3.9) * V₂
Dividing both sides by √(3.9):
V₂ = 17 V / √(3.9)
Calculating the value:
V₂ ≈ 9.40 V
Therefore, the potential difference across the plates of the capacitor filled with a dielectric should be approximately 9.40 V in order to store the same amount of electrical energy as the empty capacitor.
To solve this problem, we need to use the concept of capacitance.
The capacitance of a capacitor filled with a dielectric is given by:
C = k * Co
where C is the capacitance of the dielectric-filled capacitor, k is the dielectric constant, and Co is the capacitance of the empty capacitor.
Since the capacitors are identical, the capacitance of the empty capacitor (Co) is the same as the capacitance of the dielectric-filled capacitor (C) when k = 1.
Given that the empty capacitor is connected to a 17-V battery, we can determine the energy stored in the empty capacitor using the formula:
E = (1/2) * Co * V^2
where E is the energy stored in the capacitor and V is the potential difference across the plates.
To find the potential difference across the plates of the dielectric-filled capacitor that stores the same amount of energy, we need to equate the energy stored in both capacitors:
E = (1/2) * Co * V^2 = (1/2) * C * V^2
Since Co = C when k = 1, we can simplify the equation to:
V^2 = V^2 / k
From here, we can solve for the potential difference (V) for the dielectric-filled capacitor:
V = √(V^2 / k)
Using the given dielectric constant k = 3.9, we can substitute the values:
V = √(17^2 / 3.9) ≈ √73.44 ≈ 8.57 V
Therefore, the potential difference across the plates of the capacitor filled with a dielectric should be approximately 8.57 V to store the same amount of electrical energy as the empty capacitor connected to a 17-V battery.