Question

A parallel plate capacitor consisting of a pair of rectangular plates, each measuring \(1.00 \mathrm{~cm}\) by \(10.0 \mathrm{~cm},\) with a separation between the plates of \(0.100 \mathrm{~mm},\) is charged by a power supply at a potential difference of \(1.00 \cdot 10^{3} \mathrm{~V}\). The power supply is then removed, and without being discharged, the capacitor is placed in a vertical position over a container holding de-ionized water, with the short sides of the plates in contact with the water, as shown in the figure. Using energy considerations, show that the water will rise between the plates. Neglecting other effects, determine the system of equations that can be used to calculate the height to which the water rises between the plates. You do not have to solve the system.

Short answer

Question: Calculate the height to which the water rises between the parallel plates of a charged capacitor when it is placed vertically over a container with de-ionized water. Answer: To find the height to which the water rises between the plates, use the calculated system of equations: \(h^2 = \frac{4.43 \times 10^{-5} J}{\rho gA}\)

Step-by-step solution

Find the energy stored in the charged capacitor

The energy stored in a charged capacitor can be found using the formula: \(U = \frac{1}{2}CV^2\) where \(U\) is the stored energy, \(C\) is the capacitance of the capacitor, and \(V\) is the potential difference across the capacitor. The capacitance \(C\) of the parallel plate capacitor can be calculated using the formula: \(C = \frac{\epsilon_0 A}{d}\) where \(\epsilon_0\) is the vacuum permittivity with a value of \(8.85 \times 10^{-12} F/m\), \(A\) is the area of each plate, and \(d\) is the separation between the plates. Now, we can calculate the capacitance: \(A = 1.00 cm \times 10.0 cm = 1.00 \times 10^{-2} m \times 1.00 \times 10^{-1} m = 1.00 \times 10^{-3} m^2\) \(d = 0.100 mm = 1.00 \times 10^{-4} m\) \(C = \frac{8.85 \times 10^{-12} F/m \times 1.00 \times 10^{-3} m^2}{1.00 \times 10^{-4} m} = 8.85 \times 10^{-11} F\) Now, we can calculate the energy stored in the capacitor: \(U = \frac{1}{2}(8.85 \times 10^{-11} F)(1.00 \cdot 10^{3} V)^2 = 4.43 \times 10^{-5} J\)

Calculate the energy required for the water to rise between the plates

To find the energy required for the water to rise between the plates, we will consider the work done against gravity. The potential energy required to raise a mass of water \(m\) to a height \(h\) is given by: \(U_{water} = mgh\) where \(g\) is the acceleration due to gravity, approximately \(9.81 m/s^2\). The mass of the water can be found using the formula: \(m = \rho V\) where \(\rho\) is the density of water (approximately \(1,000 kg/m^3\)) and \(V\) is the volume of water risen between the plates. The volume of the water can be found using the formula: \(V = Ah\) where \(A\) is the area of the water's cross-section and \(h\) is the height of the water between the plates. Now, we can express \(U_{water}\) in terms of \(h\): \(U_{water} = \rho g(Ah)h\)

Equate the energies and find the system of equations

Finally, we equate the energy stored in the capacitor with the energy required for the water to rise between the plates: \(4.43 \times 10^{-5} J = \rho g(Ah)h\) We can rearrange this equation and we get the equation that can be used to calculate the height to which the water rises between the plates: \(h^2 = \frac{4.43 \times 10^{-5} J}{\rho gA}\) Thus, the system of equation needed to find the height \(h\) to which the water rises between the plates is: \(h^2 = \frac{4.43 \times 10^{-5} J}{\rho gA}\)

Key concepts

Capacitance Calculation

The capacitance of a capacitor is a measure of its ability to store charge per unit voltage increase across its plates. For a parallel plate capacitor, which is common in many physics and electronics applications, the capacitance can be calculated with the formula:

capacitance}, the associated formula would be:
\[C = \frac{\epsilon_0 A}{d}\]
where \(\epsilon_0\) represents the vacuum permittivity (approximately \(8.85 \times 10^{-12} F/m\)), \(A\) is the area of one of the plates, and \(d\) is the distance between the two plates. It's critical to ensure that all measurements are in units compatible with Farads when calculating capacitance—in this case, meters. To help students better understand, using everyday objects or simple diagrams to visually represent the concept can greatly enhance comprehension.

Energy Stored in a Capacitor

The energy stored in a capacitor is the work that has been done to move charge from one plate to another against the electric field between them. It's essentially the electrical potential energy of the system and can be calculated using the following formula:

\[U = \frac{1}{2}CV^2\]
where \(U\) is the energy in joules, \(C\) is the capacitance, and \(V\) is the potential difference (voltage) across the plates. Breaking it down further, the \(\frac{1}{2}\) factor is due to the linear relationship between voltage and charge for a capacitor. The energy calculation is instrumental in understanding not just static scenarios but dynamic ones as well, such as the filling of the capacitor in circuits or, as in our problem, the conversion of that energy to move water against gravity.

Electrostatics in Fluids

Electrostatic Attraction and Fluid Elevation

When dealing with electrostatics in fluids, especially in non-conductive (dielectric) fluids like deionized water, unique phenomena can occur. The charged plates of a capacitor can induce a dipole moment in the fluid molecules, causing a slight alignment with the electric field. This can create an attractive force that can pull the fluid between the plates.

Energy Considerations in Fluid Elevation

The energy stored in a capacitor can be converted into mechanical work, such as raising a fluid's level between the plates. To calculate the height to which the fluid rises, equating the electrical energy to the gravitational potential energy the fluid gains is essential, leading to the equation mentioned in the step-by-step solution. By taking into account the density of the fluid and the force of gravity, and understanding that energy is conserved, students can begin to appreciate the direct application of electrostatic principles in a fluid context—showing how phenomena at the microscopic level can lead to macroscopic effects.