Question

A 4.0 -nF parallel plate capacitor with a sheet of Mylar \((\kappa=3.1)\) filling the space between the plates is charged to a potential difference of \(120 \mathrm{~V}\) and is then disconnected. a) How much work is required to completely remove the sheet of Mylar from the space between the two plates? b) What is the potential difference between the plates of the capacitor once the Mylar is completely removed?

Short answer

And what is the potential difference once the Mylar is removed? Answer: a) The work required to remove the dielectric can be found using the following steps: 1. Calculate the initial capacitance of the capacitor with the dielectric, \(C = C_0 \kappa\) 2. Calculate the initial charge on the capacitor, \(Q = CV\) 3. Calculate the initial energy stored in the capacitor, \(U = \frac{1}{2}CV^2\) 4. Calculate the final capacitance without the dielectric, which will be equal to \(C_0\) 5. Calculate the final potential difference, \(V_f = \frac{Q}{C_0}\) 6. Calculate the final energy stored in the capacitor without the dielectric, \(U_f = \frac{1}{2}C_0V_f^2\) 7. Calculate the work required to remove the dielectric, \(W = U_f - U\) b) The potential difference once the Mylar is removed can be found in step 5, \(V_f = \frac{Q}{C_0}\).

Step-by-step solution

Find the initial capacitance with the dielectric

To find the initial capacitance of the capacitor, we can use the formula \(C = C_0 \kappa\), where \(C_0\) is the capacitance without the dielectric and \(\kappa\) is the dielectric constant of the Mylar. Using the given values: \(C_0 = 4.0 \times 10^{-9} ~F\) \(\kappa = 3.1\) The initial capacitance with the dielectric can thus be calculated as: \(C = C_0 \kappa\)

Calculate the initial charge on the capacitor

The initial charge (\(Q\)) on the capacitor can be calculated using the formula \(Q = C V\), where \(V\) is the potential difference. Given the potential difference as \(120 \mathrm{~V}\) and the capacitance calculated in step 1, we have: \(Q = CV\)

Calculate the initial energy stored in the capacitor

The initial energy stored in the capacitor with the dielectric can be calculated using the formula \(U = \frac{1}{2}CV^2\). Using the values of \(C\) and \(V\) from previous steps, we can find the initial energy as: \(U = \frac{1}{2}CV^2\)

Calculate the final capacitance without the dielectric

After removing the dielectric, the capacitance will be equal to the initial capacitance without the dielectric, which is \(C_0 = 4.0 \times 10^{-9} ~F\).

Calculate the final potential difference

The final potential difference (\(V_f\)) can be found using the same formula as in step 2, but now using the final capacitance without the dielectric (\(C_0\)) and solving for \(V_f\). With the charge being conserved as the capacitor is disconnected: \(V_f = \frac{Q}{C_0}\)

Calculate the final energy stored in the capacitor without the dielectric

The final energy stored in the capacitor without the dielectric can be calculated using the formula \(U_f = \frac{1}{2}C_0V_f^2\). Using the values of \(C_0\) and \(V_f\) from previous steps, we can find the final energy as: \(U_f = \frac{1}{2}C_0V_f^2\)

Calculate the work required to remove the dielectric

The work required to remove the dielectric can be calculated as the difference between the final energy and the initial energy stored in the capacitor: \(W = U_f - U\)

Finalize the results

Calculate the numerical values for each step and find the answers for part a) and part b). The work required to remove the dielectric (solution to part a)) can be directly extracted from the result in step 7, and the final potential difference (solution to part b)) from the result in step 5.