Problem 53 - University Physics with Modern Physics

Question

A 20.0 -\muF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged \(10.0-\mu \mathrm{F}\) capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

Step-by-Step Solution

Answer

(a) 16 mC, (b) 533.33 V, (c) 0.0427 J, (d) decreased by 0.0213 J.

Step 1: Calculate the Original Charge

To find the original charge on the 20 \( \mu \)F capacitor, use the formula for the charge \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the potential difference. Substitute \( C = 20 \times 10^{-6} \mathrm{F} \) and \( V = 800 \mathrm{V} \):\[ Q = 20 \times 10^{-6} \times 800 = 16 \times 10^{-3} \mathrm{C} \]Thus, the original charge \( Q \) is 16 mC.

Step 2: Determine Total Capacitance After Connection

When the charged 20 \( \mu \)F capacitor is connected to the uncharged 10 \( \mu \)F capacitor, they are in parallel, and the total capacitance \( C_{total} \) is the sum of the two:\[ C_{total} = 20 + 10 = 30 \ \mu \text{F} \]\.

Step 3: Calculate Final Potential Difference

The charge is conserved. Thus, the charge in the system remains 16 mC. Use the formula \( V_f = \frac{Q}{C_{total}} \) to find the final potential difference. Substitute \( Q = 16 \times 10^{-3} \) C and \( C_{total} = 30 \times 10^{-6} \) F:\[ V_f = \frac{16 \times 10^{-3}}{30 \times 10^{-6}} = 533.33 \mathrm{V} \]The final potential difference across each capacitor is approximately 533.33 V.

Step 4: Calculate Final Energy of the System

The final energy in the system is calculated using the formula for the energy stored in capacitors: \( U = \frac{1}{2} C V^2 \). With \( C_{total} = 30 \times 10^{-6} \) F and \( V_f = 533.33 \) V, compute:\[ U_f = \frac{1}{2} \times 30 \times 10^{-6} \times (533.33)^2 \approx 4.27 \times 10^{-2} \mathrm{J} \]

Step 5: Calculate Decrease in Energy

The initial energy stored in the 20 \( \mu \)F capacitor before connection is:\[ U_i = \frac{1}{2} \times 20 \times 10^{-6} \times (800)^2 = 6.4 \times 10^{-2} \mathrm{J} \]The decrease in energy when the capacitors are connected is:\[ \Delta U = U_i - U_f = 6.4 \times 10^{-2} - 4.27 \times 10^{-2} = 2.13 \times 10^{-2} \mathrm{J} \]

Key Concepts

Charge Conservation

In the realm of capacitance, charge conservation is a fundamental principle that holds everything together. It states that the total electric charge in an isolated system remains constant. This means that when capacitors are connected in a system, like our charged 20 \( \mu \mathrm{F} \) capacitor connecting to an uncharged 10 \( \mu \mathrm{F} \) capacitor, the charge does not magically appear or disappear.
Instead, the charge redistributes among the capacitors. Initially, the charged capacitor holds all the charge. Once the connection is made, the charge spreads across the capacitors but the total amount remains the same. Even when potential differences change, the conserved charge is simply shuffled.

• Original Charge: 16 mC before connection
• Charge After Connection: Remains 16 mC since it is conserved

This preserves the charge balance, ensuring our calculations stay true to nature's laws.

Capacitors in Parallel

Capacitors can be arranged in several configurations, with parallel being one of the most straightforward. When capacitors are connected in parallel, the total capacitance becomes the sum of their individual capacitances.
In our exercise, when the charged 20 \( \mu \mathrm{F} \) capacitor is connected to the uncharged 10 \( \mu \mathrm{F} \) capacitor, they form a parallel circuit.

• Total Capacitance: \( C_{total} = C_1 + C_2 = 20 + 10 = 30 \ \, \mu \mathrm{F} \)

This addition means both capacitors experience the same voltage drop once they are connected, which simplifies the calculations for energy and potential difference. Capacitors in parallel combine capacity, allowing them to store more charge, yet they remain versatile and stable when distributing potential.

Energy in Capacitors

In physics, energy is ubiquitous and capacitors have their way of storing it as well. The energy stored in a capacitor is given by the formula \( U = \frac{1}{2} C V^2 \). This tells us that energy depends on both the capacitance of the capacitor and the squared potential difference across it.
For the 20 \( \mu \mathrm{F} \) capacitor charged to 800 V, the initial energy is 0.064 J. However, once connected in parallel with the uncharged capacitor, the system's energy differs:

• Final Energy in the System: \( U_f = \frac{1}{2} \times 30 \times 10^{-6} \times (533.33)^2 = 0.0427 \, \mathrm{J} \)
• Energy Decrease: \( \Delta U = 0.064 - 0.0427 = 0.0213 \, \mathrm{J} \)

Energy reduction happens because while charge is conserved, some energy is lost as heat or other forms of energy due to the resistance of the connecting wires and other inevitable losses.

Potential Difference Across Capacitors

Potential difference, or voltage, across capacitors is key because it dictates how charges distribute and how energy is stored in the system. Initially, our 20 \( \mu \mathrm{F} \) capacitor is charged to an impressive 800 V.
But, upon connecting to a 10 \( \mu \mathrm{F} \) uncharged capacitor in parallel, this potential difference changes, equalizing due to the shared system:

• Final Potential Difference Across Each Capacitor: \( V_f = \frac{Q}{C_{total}} = \frac{16 \, \text{mC}}{30 \times 10^{-6} \mathrm{F}} = 533.33 \, \text{V} \)

This uniform potential difference ensures that once combined, both capacitors share the electric field effect equally. The adjusting of potential difference is a natural result of ensuring that the charge is universally balanced across the capacitors.