When an air capacitor with a capacitance of 400 (1 = ) is connected to a power supply, the energy stored in the capacitor is 2.05×10−5 . 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 2.28×10−5

Question

When an air capacitor with a capacitance of 400 (1 = ) is connected to a power supply, the energy stored in the capacitor is 2.05×10−5 . 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 2.28×10−5

Answer 1

And the question is....

Answer 2

To solve this problem, we can use the formula for the energy stored in a capacitor:

Energy = (1/2) * C * V^2

Where:
- Energy is the stored energy in the capacitor,
- C is the capacitance of the capacitor,
- V is the voltage across the capacitor.

We are given the initial capacitance as 400 pF (picoFarads), which is equivalent to 400 × 10^(-12) Farads.

Let's assume the initial voltage across the capacitor is V1.
The initial energy stored in the capacitor can be calculated using the given values:

2.05 × 10^(-5) = (1/2) * 400 × 10^(-12) * V1^2

Simplifying:

2.05 × 10^(-5) = 2 × 10^(-14) * V1^2

To get the initial voltage V1, we can rearrange the equation:

V1^2 = (2.05 × 10^(-5)) / (2 × 10^(-14))

V1^2 = 1.025

Taking the square root of both sides:

V1 = √1.025

V1 ≈ 1.012 V

Now, let's consider the situation after the dielectric is inserted, where the stored energy increases to 2.28 × 10^(-5).

The capacitance with the dielectric is given by:

C' = k * C

Where:
- C' is the new capacitance with the dielectric,
- k is the dielectric constant,
- C is the initial capacitance.

Since the slab of dielectric completely fills the space between the plates, it increases the capacitance. The new energy can be calculated using this new capacitance.

2.28 × 10^(-5) = (1/2) * (k * 400 × 10^(-12)) * V2^2

Simplifying:

2.28 × 10^(-5) = (k * 2 × 10^(-14)) * V2^2

Dividing both sides by V1^2 (since V1 = V2):

(2.28 × 10^(-5)) / (V1^2) = (k * 2 × 10^(-14)) * (V2^2 / V1^2)

Simplifying further:

(2.28 × 10^(-5)) / (V1^2) = k * (2 × 10^(-14))

Finally, solving for k:

k = [(2.28 × 10^(-5)) / (V1^2)] / (2 × 10^(-14))

Substituting the value of V1 = 1.012 V:

k = [(2.28 × 10^(-5)) / (1.012^2)] / (2 × 10^(-14))

k ≈ 113.26

Therefore, the dielectric constant (k) of the inserted slab is approximately 113.26.