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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.

Short Answer

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

Step by step solution

01

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.
02

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:\[ C_{total} = 20 + 10 = 30 \ \mu \text{F} \]\.
03

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.
04

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:\[ U_f = \frac{1}{2} \times 30 \times 10^{-6} \times (533.33)^2 \approx 4.27 \times 10^{-2} \mathrm{J} \]
05

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:\[ \Delta U = U_i - U_f = 6.4 \times 10^{-2} - 4.27 \times 10^{-2} = 2.13 \times 10^{-2} \mathrm{J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A cylindrical air capacitor of length 15.0 \(\mathrm{m}\) stores \(3.20 \times 10^{-9} \mathrm{J}\) of energy when the potential difference between the two conductors is 4.00 \(\mathrm{V}\) . (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Three capacitors having capacitances of \(8.4,8.4,\) and 4.2\(\mu \mathrm{F}\) are connected in series across a \(36-\mathrm{V}\) potential difference. (a) What is the charge on the \(4.2-\mu F\) capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors?

A parallel-plate capacitor has plates with area 0.0225 \(\mathrm{m}^{2}\) separated by 1.00 \(\mathrm{mm}\) of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 \(\mathrm{V}\) . (b) Use Gauss's law (Eq. 24.23 ) to calculate the electric field inside the Teflon. (c) Use Gauss's law to calculate the electric field if the voltage source is disconnected and the Teflon is removed.

You are working on an electronics project requiring a variety of capacitors, but you have only a large supply of 100 -nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) 50 \(\mathrm{nF}\) ; (b) \(450 \mathrm{nF} ;\) (c) \(25 \mathrm{nF} ;\) (d) 75 \(\mathrm{nF}\) .

B10 Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{\boldsymbol{E}}\) in the wall-between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) Atypical cell in the human body has a volume of\(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

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