When an air capacitor with a capacitance of 300 nF (1 {\rm nF} = 10^{-9}\;{\rm F}) is connected to a power supply, the energy stored in the capacitor is 1.95×10−5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.74×10−5 J.
What is the potential difference between the capacitor plates?
What is the dielectric constant of the slab?
With the power always connected to the same supply voltage V, that remains the potential difference between the plates. The energy stored is (1/2) C V^2. C must increase by a ratio 2.74/1.95 = 1.405
That equals the diectric constant of the gap material.
Well, to find the potential difference between the capacitor plates, we can use the formula:
\(E = \frac{1}{2}CV^2\)
Where E is the energy stored in the capacitor, C is the capacitance, and V is the potential difference.
Given that the energy initially stored is 1.95×10−5 J, and the capacitance is 300 nF (1 nF = 10^-9 F), we can rearrange the formula to solve for V:
\(V = \sqrt{\frac{2E}{C}}\)
\(V = \sqrt{\frac{2 \times 1.95\times10^{-5} \, \text{J}}{300 \times 10^{-9} \, \text{F}}}\)
Now, let me put this into my super advanced calculation clown machine...
*Clown Bot presses buttons and sounds of boings and honks can be heard*
And voila! The potential difference between the capacitor plates is approximately 1.12 V.
Now, moving on to the dielectric constant of the slab. The increase in stored energy when the dielectric is inserted is given as 2.74×10−5 J. We can use another formula:
\(\Delta E = \frac{1}{2} C \, \Delta V^2 \, (1 - \frac{1}{k})\)
Where ΔE is the increase in energy, ΔV is the change in potential difference, C is the capacitance, and k is the dielectric constant.
Rearranging the formula, we can solve for k:
\(k = 1 - \frac{2 \Delta E}{C \Delta V^2}\)
Plugging in the values:
\(k = 1 - \frac{2 \times 2.74\times10^{-5} \, \text{J}}{300 \times 10^{-9} \, \text{F} \times (\text{change in potential difference})^2}\)
Unfortunately, I don't have access to the value of the change in potential difference, so I can't fully solve this for you. But once you have that value, you can put it into the formula and calculate the dielectric constant.
Remember, laughter is the best capacitor for any situation!
To find the potential difference between the capacitor plates, we can use the formula for the energy stored in a capacitor:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor
C is the capacitance of the capacitor
V is the potential difference between the capacitor plates
Let's start by finding the potential difference V based on the given energy and capacitance values.
Given:
C = 300 nF = 300 * 10^(-9) F
E = 1.95 * 10^(-5) J
Using the formula, we can rearrange it to solve for V:
V = sqrt(2 * E / C)
Substituting the given values:
V = sqrt(2 * 1.95 * 10^(-5) J / (300 * 10^(-9) F))
Simplifying:
V = sqrt(2 * 1.95 * 10^(-5) J * (1 / 300 * 10^(-9) F))
V = sqrt((2 * 1.95 * 10^(-5) J) / (300 * 10^(-9) F))
V = sqrt(2 * 1.95 * 10^(-5) J) / sqrt(300 * 10^(-9) F)
V = sqrt(2 * 1.95 * 10^(-5) J) / (sqrt(300) * sqrt(10^(-9) F))
V ≈ 61.135 V
Therefore, the potential difference between the capacitor plates is approximately 61.135 volts.
To find the dielectric constant of the slab, we need to compare the original stored energy with the increased stored energy after the slab is inserted.
Given:
Change in energy (ΔE) = 2.74 × 10^(-5) J
The change in energy is related to the change in capacitance (ΔC) and the potential difference (V) by the formula:
ΔE = (1/2) * ΔC * V^2
Since the capacitance is directly proportional to the dielectric constant (K), we can write:
ΔC = C * (K - 1)
Substituting this into the equation above, we get:
ΔE = (1/2) * C * (K - 1) * V^2
Simplifying, we have:
2 * ΔE = C * (K - 1) * V^2
Plugging in the known values:
2 * 2.74 × 10^(-5) J = (300 * 10^(-9) F) * (K - 1) * (61.135 V)^2
2.74 × 10^(-5) J = (300 * 10^(-9) F) * (K - 1) * (61.135 V)^2
Simplifying further, we find:
K - 1 = (2.74 × 10^(-5) J) / ((300 * 10^(-9) F) * (61.135 V)^2)
K - 1 ≈ 2.8286
K ≈ 3.8286
Therefore, the dielectric constant of the slab is approximately 3.8286.