Question

A parallel plate capacitor with vacuum between the plates has a capacitance of \(3.669 \mu \mathrm{F}\). A dielectric material with \(\kappa=3.533\) is placed between the plates, completely filling the volume between them. The capacitor is then connected to a battery that maintains a potential difference \(V\) across the plates. The dielectric material is pulled out of the capacitor, which requires \(7.389 \cdot 10^{-4} \mathrm{~J}\) of work. What is the potential difference, \(V ?\)

Short answer

Answer: The potential difference across the parallel plate capacitor is approximately 1.260 V.

Step-by-step solution

Calculate Capacitance C2 after inserting the dielectric

To find C2, we can use the formula C2 = κC1, where κ is the dielectric constant. C2 = (3.533)(3.669 * 10^{-6} F) = 12.954 * 10^{-6} F

Find the initial and final energies stored in the capacitor

We can use the formula U = (1/2) * C * V^2 to find the energy stored before and after inserting the dielectric material. Also, we know that the potential difference V does not change when a dielectric is inserted. U1 = (1/2) * C1 * V^2 U2 = (1/2) * C2 * V^2

Calculate the change in energy

The difference in energy between the initial and final states (ΔU) is equal to the work required to remove the dielectric (W). ΔU = U2 - U1 = W

Express the potential difference V in terms of given values

We can express the potential difference V by substituting the expressions for U1 and U2 from Step 2 and ΔU from Step 3. (1/2) * C2 * V^2 - (1/2) * C1 * V^2 = W (1/2) * (C2 - C1) * V^2 = W V^2 = (2 * W) / (C2 - C1)

Find the potential difference V

Now, we can substitute the values for W, C1, and C2 in the expression for V^2, and finally find V. V^2 = 2 * (7.389 * 10^{-4} J) / (12.954 * 10^{-6} F - 3.669 * 10^{-6} F) V^2 = (2 * 7.389 * 10^{-4}) / (9.285 * 10^{-6}) V = sqrt(1.587443) V ≈ 1.260 V The potential difference across the parallel plate capacitor is approximately 1.260 V.