Two capacitors are identical, except that one is empty and the other is filled with a dielectric (k = 4.2). The empty capacitor is connected to a 14 -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?
Capacitance of 1 = C
Capacitance of 2 = k*C
Energy = (1/2)*C*V^2
As per the question,
(1/2)*(C1)*(V1)^2 = (1/2)*(C2)*(V2)^2
=> C*(14)^2 = k*C*(V2)^2
Solve for V2
Well, well, well, we have a shocking situation here! So, we have an empty capacitor and another one filled with a dielectric material. Let's call the potential difference across the plates of the filled capacitor V (we don't want to hurt its feelings by calling it "empty," too).
Now, when it comes to storing electrical energy, it's all about the capacitance, my electrically curious friend. Since both capacitors are identical, their capacitance (C) is also the same. The capacitance is determined by the equation:
C = (k * ε₀ * A) / d,
where k is the dielectric constant, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the separation of the plates.
Okay, now let's get back to business. The empty capacitor is connected to a 14-V battery. We need to find the potential difference across the plates of the filled capacitor so that it stores the same amount of electrical energy as the empty one.
Since the energy stored in a capacitor is given by the equation:
E = (1/2) * C * V²,
we want the energy to be the same for both capacitors. But here's the catch: the capacitance of the filled capacitor is different due to the dielectric material.
So, to make the energy the same for both capacitors, we need to adjust the potential difference (V) across the plates of the filled capacitor. How exactly? Well, by taking into account the different capacitance caused by the dielectric material!
Using the equation for the energy stored in a capacitor, we can set up an equation:
(1/2) * C₁ * V₁² = (1/2) * C₂ * V₂²,
where C₁ and V₁ represent the capacitance and potential difference of the empty capacitor, and C₂ and V₂ represent the capacitance and potential difference of the filled capacitor.
Since we know that C₂ = k * C₁ (due to the dielectric constant), we can rewrite the equation as:
(1/2) * C₁ * V₁² = (1/2) * (k * C₁) * V₂².
After a bit of rearranging and canceling, we find:
V₂² = (V₁² / k).
Now, we can plug in the values we know:
V₂² = (14² / 4.2).
Calculating that, we find:
V₂ ≈ 33.33 V.
So, to store the same amount of electrical energy, the potential difference across the plates of the filled capacitor should be approximately 33.33 volts.
Now, that's a true electrifying balancing act!
To find the potential difference across the plates of the capacitor filled with a dielectric, let's denote the capacitance of the empty capacitor as C0 and the capacitance of the capacitor with a dielectric as Cd.
The electrical energy stored in a capacitor is given by the formula:
E = (1/2) * C * V^2
Where E is the electrical energy, C is the capacitance, and V is the potential difference across the plates.
Since we want the two capacitors to store the same amount of electrical energy, we can set up the following equation:
(1/2) * C0 * V0^2 = (1/2) * Cd * Vd^2
Where V0 is the potential difference across the plates of the empty capacitor and Vd is the potential difference across the plates of the capacitor with a dielectric.
The capacitance of a capacitor is given by the formula:
C = (k * ε0 * A)/d
Where k is the dielectric constant, ε0 is the permittivity of free space, A is the area of the capacitor plates, and d is the distance between the plates.
Since the two capacitors are identical except for the dielectric, their areas and distances are the same. Hence, we can write:
C0 = Cd = (k * ε0 * A)/d
Since we know the value of k (k = 4.2), we can substitute this value into the equation:
C0 = Cd = (4.2 * ε0 * A)/d
Now we can rearrange the initial equation and substitute the value above:
(1/2) * C0 * V0^2 = (1/2) * Cd * Vd^2
(1/2) * [(4.2 * ε0 * A)/d] * V0^2 = (1/2) * [(4.2 * ε0 * A)/d] * Vd^2
Simplifying further:
V0^2 = Vd^2
Taking the square root of both sides:
V0 = Vd
Therefore, the potential difference across the plates of the capacitor filled with a dielectric must be the same as the potential difference across the plates of the empty capacitor in order for them to store the same amount of electrical energy.
To solve this problem, we need to understand the relationship between the capacitance, voltage, and stored energy in a capacitor.
The capacitance of a capacitor is given by the formula C = ε₀ * A / d, where C is the capacitance, ε₀ is the vacuum permittivity, A is the area of the plates, and d is the distance between the plates.
The voltage across a capacitor is given by V = Q / C, where V is the voltage, Q is the charge stored on the plates, and C is the capacitance.
The stored energy in a capacitor is given by U = 1/2 * C * V², where U is the stored energy, C is the capacitance, and V is the voltage.
Since the two capacitors are identical in all respects except for the dielectric, the capacitance of both capacitors will be the same.
Let's denote the capacitance of both capacitors as C.
For the empty capacitor:
C₁ = C
For the capacitor filled with the dielectric:
C₂ = k * C
where k is the dielectric constant.
Since the electrical energy stored in the capacitors is the same, we can equate their energies:
1/2 * C₁ * V₁² = 1/2 * C₂ * V₂²
Substituting the values of C₁, C₂, and k into the equation, we get:
1/2 * C * V₁² = 1/2 * (k * C) * V₂²
Simplifying the equation, we find:
V₁² = k * V₂²
Since we know the voltage V₁ (14 V) of the empty capacitor, we can solve for V₂:
14² = k * V₂²
Taking the square root of both sides, we get:
V₂ = sqrt(14² / k)
Substituting the value of k (4.2) into the equation, we finally get the potential difference across the plates of the capacitor filled with the dielectric:
V₂ = sqrt(14² / 4.2) ≈ 7.07 V
Therefore, the potential difference across the plates of the capacitor filled with a dielectric should be approximately 7.07 V in order to store the same amount of electrical energy as the empty capacitor.