Question

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and has area \(A=100 . \mathrm{cm}^{2}\) and separation distance \(d=2.50 \mathrm{~cm} .\) a) Find the capacitance, \(C\), the potential difference, \(V\), the electric field, \(E\), the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

Short answer

Question: Calculate the capacitance, potential difference, electric field, charge, and potential energy of a parallel plate capacitor before and after a dielectric slab is inserted, when the battery is connected during insertion and when it's disconnected. Answer: a) Capacitor in air (no dielectric material): 1. Capacitance: \(3.54 \times 10^{-12} F\) 2. Potential difference: \(110 V\) 3. Electric field: \(4400 V/m\) 4. Charge: \(3.89 \times 10^{-10} C\) 5. Potential energy: \(2.14 \times 10^{-8} J\) b) Capacitor with dielectric material and battery connected: 1. New capacitance: \(8.18 \times 10^{-12} F\) 2. Electric field: \(1904.76 V/m\) 3. Charge: \(9.00 \times 10^{-10} C\) 4. Potential energy: \(4.93 \times 10^{-8} J\) c) Capacitor with dielectric material and battery disconnected: 1. Capacitance: \(8.18 \times 10^{-12} F\) 2. New potential difference: \(1100 V\) 3. Electric field: \(44000 V/m\) 4. Charge: \(9.00 \times 10^{-10} C\) 5. Potential energy: \(4.93 \times 10^{-7} J\)

Step-by-step solution

Find the capacitance, C

The capacitance of a parallel plate air capacitor is given by the formula \(C=\frac{A\varepsilon_0}{d}\), where \(\varepsilon_0\) is the vacuum permittivity \((\varepsilon_0 \approx 8.854 \times 10^{-12} F/m)\), \(A\) is the plate area, and \(d\) is the distance between the plates. Use the given values to calculate the capacitance as follows: $$C=\frac{A\varepsilon_0}{d}=\frac{1.00\times 10^{-2} \cdot 8.854\times 10^{-12} }{2.50 \times 10^{-2}} = 3.54 \times 10^{-12} \mathrm{F}$$

Find the potential difference, V

The potential difference is already given in the problem statement: \(V = 110\;\text{V}\).

Find the electric field, E

The electric field in a parallel plate capacitor can be calculated using the formula \(E = \frac{V}{d}\). Using the given values and the calculated capacitance, find the electric field: $$E = \frac{V}{d} = \frac{110}{2.50 \times 10^{-2}} = 4400\;\text{V/m}$$

Find the total charge stored on the capacitor, Q

The charge stored on a capacitor can be calculated using the formula \(Q = CV\). Using the calculated capacitance and the given potential difference, find the charge stored on the capacitor: $$Q = CV = 3.54 \times 10^{-12} \cdot 110 = 3.89 \times 10^{-10}\; \text{C}$$

Find the electric potential energy stored in the capacitor, U

The electric potential energy stored in a capacitor can be found using the formula \(U = \frac{1}{2}CV^2\). Using the calculated capacitance and the given potential difference, find the electric potential energy stored in the capacitor: $$U = \frac{1}{2}CV^2 = \frac{1}{2}(3.54\times 10^{-12})(110^2) = 2.14 \times 10^{-8}\;\text{J}$$ b) Capacitor with dielectric material and battery connected: || - || -

Find the new capacitance, C

When dielectric material is inserted, the new capacitance can be found using the formula \(C' = \kappa C\), where \(\kappa\) is the dielectric constant. Use the calculated capacitance from part (a) and the given dielectric constant to find the new capacitance: $$C' = \kappa C = 2.31(3.54\times 10^{-12}) = 8.18 \times 10^{-12}\; \text{F}$$ (Note that the potential difference, V, remains the same since the battery is still connected.)

Find the new electric field, E

The electric field in a capacitor with a dielectric can be found using the formula \(E' = \frac{E}{\kappa}\). Find the new electric field using the calculated electric field from part (a) and the given dielectric constant: $$E' = \frac{E}{\kappa} = \frac{4400}{2.31} = 1904.76\;\text{V/m}$$

Find the new total charge stored on the capacitor, Q

The new charge stored on the capacitor can be calculated using the formula \(Q' = C'V\). Using the new calculated capacitance and the given potential difference, find the new charge stored: $$Q' = C'V = 8.18 \times 10^{-12} \cdot 110 = 9.00 \times 10^{-10}\; \text{C}$$

Find the new electric potential energy stored in the capacitor, U

The new electric potential energy stored in the capacitor can be found using the formula \(U' = \frac{1}{2}C'V^2\). Using the new calculated capacitance and the given potential difference, find the new electric potential energy stored in the capacitor: $$U' = \frac{1}{2}C'V^2 = \frac{1}{2}(8.18\times 10^{-12})(110^2) = 4.93 \times 10^{-8}\;\text{J}$$ c) Capacitor with dielectric material and battery disconnected || - || -

Find the new capacitance, C

The capacitance remains the same as in part (b) since the dielectric slab is still inserted. So, \(C'= 8.18 \times 10^{-12}\;\text{F}\).

Find the new potential difference, V

When the battery is disconnected, the charge stored on the capacitor remains the same. To find the new potential difference, use the formula \(V' = \frac{Q}{C'}\). Using the new charge value \(Q'=9.00 \times 10^{-10}\;\text{C}\) from part (b) and the capacitance value \(C'=8.18 \times 10^{-12}\;\text{F}\), find the new potential difference: $$V' = \frac{Q'}{C'} = \frac{9.00 \times 10^{-10}}{8.18\times 10^{-12}} = 1100\;\text{V}$$

Find the new electric field, E

The electric field in a capacitor with a dielectric can be found using the formula \(E' = \frac{V'}{d}\). Find the new electric field using the new calculated potential difference and the separation distance: $$E' = \frac{V'}{d} = \frac{1100}{2.50 \times 10^{-2}} = 44000\;\text{V/m}$$

Find the new total charge stored on the capacitor, Q

The total charge remains the same as in part (b) since the battery is disconnected, which means that \(Q'=9.00 \times 10^{-10}\;\text{C}\).

Find the new electric potential energy stored in the capacitor, U

The new electric potential energy stored in the capacitor can be found using the formula \(U' = \frac{1}{2}C'{V'}^2\). Using the new calculated capacitance and the new potential difference, find the new electric potential energy stored in the capacitor: $$U' = \frac{1}{2}C'{V'}^2 = \frac{1}{2}(8.18\times 10^{-12})(1100^2) = 4.93 \times 10^{-7}\;\text{J}$$