Question

When an air capacitor with a capacitance of 330 nF (1 nF = 10⁻⁵ F) is connected to a power supply, the energy stored in the capacitor is 2.25×10−5J. 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 2.42×10⁻⁵J. a)What is the potential difference between the capacitor plates? b)What is the dielectric constant of the slab?

Answer 1

The potential difference between the capacitor plates can be calculated using the relationship between stored energy and capacitance. Subsequently, with the increase in stored energy due to the insertion of the dielectric, the dielectric constant can be determined utilizing the fact that the potential difference remains constant while the capacitor is connected to the power supply.

Step-by-step explanation:

To calculate the potential difference (V) between the plates of the capacitor, we can use the energy stored in the capacitor, which is given by the formula:

E = 1/2 * C * V^2

Where E is the energy (in joules), C is the capacitance (in farads), and V is the potential difference across the plates of the capacitor. Using the given energy (2.25×10⁻µJ) and the capacitance (330 nF), we can rearrange the formula to solve for V:

V = √(2E/C)

After calculating V, we'll proceed to find the dielectric constant (K) of the slab. The new energy stored, with the dielectric inserted, is the original energy plus the increase (2.25×10⁻µJ + 2.42×10⁻µJ). Since the potential difference remains the same when the capacitor is connected to the power supply, the capacitance must have increased because of the dielectric insertion. We can find K with the following steps:

New capacitance, C' = C * K

New energy, E' = 1/2 * C' * V^2

We can set up an equation with the known and new energy values and solve for K:

K = E' / E

Since C doesn't change, it cancels out in the equation and allows us to solve directly for K.