A parallel plate capacitor with air in the gap between the plates is connected to a \(6.00-\mathrm{V}\) battery. After charging, the energy stored in the capacitor is \(72.0 \mathrm{~nJ}\). Without disconnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional \(317 \mathrm{~nJ}\) of energy flows from the battery to the capacitor. a) What is the dielectric constant of the dielectric? b) If each of the plates has an area of \(50.0 \mathrm{~cm}^{2}\), what is the charge on the positive plate of the capacitor after the dielectric has been inserted? c) What is the magnitude of the electric field between the plates before the dielectric is inserted? d) What is the magnitude of the electric field between the plates after the dielectric is inserted?

Step by step solution

01

a) Finding the dielectric constant

The total energy stored in the capacitor after the dielectric is inserted is \(389 \mathrm{~nJ} (72.0 \mathrm{~nJ} + 317 \mathrm{~nJ})\). The energy in a capacitor is given by the equation \(E = \frac{1}{2}CV^2\), where C is the capacitance and V is the voltage. As the capacitor was not disconnected, V remains the same. But beause the capacitor got additional energy, the capacitance must have increased: \(C = C_{0}k\), where \(C_0\) is the initial capacitance (with air), and k is the dielectric constant we want to find. Therefore: \(389\mathrm{~nJ} = \frac{1}{2}C_{0}k(6.00\mathrm{V})^2\) and \(72.0\mathrm{~nJ} = \frac{1}{2}C_{0}(6.00\mathrm{~V})^2\) Divide the first equation by the second equation: \(k = \frac{389\mathrm{~nJ}}{72.0\mathrm{~nJ}} = 5.4\) The dielectric constant of the dielectric is 5.4.

02

b) Finding the charge on the positive plate

To find the charge, first we need to calculate the initial capacitance with air, \(C_0\), from the initial energy: \(C_0 = \frac{2 \times 72.0\mathrm{~nJ}}{(6.00\mathrm{~V})^2} = 4.00 \mathrm{~nF}\) Now let's find the capacitance with the dielectric, \(C = C_0 k = 4.00\mathrm{~nF} \times 5.4 = 21.6\mathrm{~nF}\). Finally, we can find the charge on the positive plate after the dielectric is inserted using the equation \(Q = CV\). So, \(Q = 21.6\mathrm{~nF} \times 6.00\mathrm{~V} = 129.6\mathrm{~nC}\). The charge on the positive plate after the dielectric has been inserted is \(129.6\mathrm{~nC}\).

03

c) Electric field before the dielectric is inserted

To find the magnitude of the electric field, first we need to calculate the distance between the plates given the initial capacitance \(C_0\) and plate area, \(A = 50.0\mathrm{~cm}^2 = 0.0050\mathrm{~m}^2\). The capacitance is given by \(C_0 = \frac{\epsilon _0A}{d}\), where \(\epsilon _0 = 8.85 \times 10^{-12} \mathrm{F/m}\) and d is the distance between the plates. Solving for d: \(d = \frac{\epsilon _0A}{C_0} = \frac{8.85 \times 10^{-12} \mathrm{F/m} \times 0.0050\mathrm{~m}^2}{4.00 \mathrm{~nF}} = 1.11 \times 10^{-3}\mathrm{~m}\). Now we can find the initial electric field, \(E_0 = \frac{V}{d} = \frac{6.00\mathrm{~V}}{1.11 \times 10^{-3}\mathrm{~m}} = 5.41 \times 10^3 \mathrm{V/m}\). The magnitude of the electric field between the plates before the dielectric is inserted is \(5.41 \times 10^3 \mathrm{V/m}\).

04

d) Electric field after the dielectric is inserted

The electric field between the plates after the dielectric is inserted is given by \(E = \frac{E_0}{k} = \frac{5.41 \times 10^3 \mathrm{V/m}}{5.4} = 1.00 \times 10^3 \mathrm{V/m}\). The magnitude of the electric field between the plates after the dielectric is inserted is \(1.00 \times 10^3 \mathrm{V/m}\).

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