When an air capacitor with a capacitance of 380 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 2.62×10−5 J.
A) What is the potential difference between the capacitor plates?
B)What is the dielectric constant of the slab?
Stored Energy = (1/2)CV^2
It increases by a factor (2.62+1.95)/1.95 = 2.344
A) Since V is constant, C increases by the same factor, to
2.344*380 nF = 891 nF
B) The dielectric constant is the factor that increased the capacitance, 2.344
To solve this problem, we can use the equation for the energy stored in a capacitor:
E = (1/2) * C * V²
where E is the energy stored in the capacitor, C is the capacitance, and V is the potential difference between the plates.
Let's use this equation to solve part A:
Given:
C = 380 nF = 380 × 10^(-9) F
E = 1.95 × 10^(-5) J
Using the equation for energy stored in a capacitor:
1.95 × 10^(-5) J = (1/2) * (380 × 10^(-9) F) * V²
Simplifying the equation:
3.9 × 10^(-5) J = (380/2) × 10^(-9) F * V²
Dividing both sides of the equation by (380/2) × 10^(-9) F:
V² = (3.9 × 10^(-5) J) / ((380/2) × 10^(-9) F)
V² = (3.9 × 10^(-5) J) / (190 × 10^(-9) F)
V² = (3.9 / 190) × 10^(-5 - (-9)) J/F
V² = (3.9 / 190) × 10^4 V
V² ≈ 0.020526 V²
Taking the square root of both sides:
V ≈ √(0.020526 V²)
V ≈ 0.143 V
Therefore, the potential difference between the capacitor plates is approximately 0.143 V.
Now let's move on to part B:
Given:
E' = 2.62 × 10^(-5) J (increase in stored energy when the dielectric is inserted)
We can again use the equation for energy stored in a capacitor:
2.62 × 10^(-5) J = (1/2) * C * V'^2
where V' is the potential difference when the dielectric is inserted.
Since the dielectric constant (k) is the ratio of the potential differences V' to V:
k = V' / V
Rearranging the equation:
V' = k * V
Substituting the given values:
2.62 × 10^(-5) J = (1/2) * (380 × 10^(-9) F) * (kV)²
Dividing both sides by (1/2) * (380 × 10^(-9) F):
5.24 × 10^(-5) J/F = (kV)²
Taking the square root of both sides:
√(5.24 × 10^(-5) J/F) = kV
√(5.24 × 10^(-5)) / √F = kV
√(5.24 × 10^(-5)) / (√1 × 10^(-9)) = kV
√5.24 / 10^(-2.5) = kV
√5.24 / 3.162 × 10^(-2) = kV
√(5.24 / 3.162) × 10^2 = kV
√1.656 × 10^2 = kV
1.286 × 10 = kV
k ≈ 1.286 × 10
Therefore, the dielectric constant of the slab is approximately 1.286 × 10.