/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A \(600 \mathrm{pF}\) capacitor ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 25

A \(600 \mathrm{pF}\) capacitor is charged by a \(200 \mathrm{~V}\) supply. It is then disconnected from the supply and is connected to another uncharged \(600 \mathrm{pF}\) capacitor. What is the common potential (in \(\mathrm{V}\) ) and energy lost (in J) after reconnection? a. \(100,6 \times 10^{-6}\) b. \(200,6 \times 10^{-5}\) c. \(200,5 \times 10^{-6}\) d. \(100,6 \times 10^{-5}\)

Short Answer

Expert verified
The common potential is 100 V and the energy lost is \(6 \times 10^{-6}\) J.

Step by step solution

01

Calculate Initial Charge on Capacitor

When the 600 pF capacitor is charged to 200 V, the charge \( Q \) on the capacitor is given by the formula \( Q = C \times V \), where \( C \) is the capacitance and \( V \) is the voltage. Plugging in the values, \( Q = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-8} \ ext{Coulombs} \).
02

Calculate Common Voltage After Connection

When the initial capacitor is connected to the uncharged capacitor of the same capacitance, the charge distributes equally among them. The total capacitance now is \( C_{total} = 600 + 600 = 1200 \, \text{pF} \). The common voltage \( V_{common} \) is given by \( V_{common} = \frac{Q_{total}}{C_{total}} = \frac{1.2 \times 10^{-8}}{1200 \times 10^{-12}} = 100 \, \text{V} \).
03

Calculate Initial Energy Stored

Initial energy stored in the capacitor is given by \( E_{initial} = \frac{1}{2} \times C \times V^2 \). Substituting the known values, \( E_{initial} = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \, \text{J} \).
04

Calculate Final Energy Stored

After the capacitors are connected, both store energy. The total stored energy after connection is \( E_{final} = \frac{1}{2} \times C_{total} \times V_{common}^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times 100^2 = 6 \times 10^{-6} \, \text{J} \).
05

Calculate Energy Lost

The energy lost due to redistribution of charge is the difference between the initial and final energies. \( E_{lost} = E_{initial} - E_{final} = 1.2 \times 10^{-5} - 6 \times 10^{-6} = 6 \times 10^{-6} \, \text{J} \).

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Key Concepts

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

Charge Distribution
When we consider capacitors, understanding how charge distribution works is key. Initially, a charged capacitor holds a certain amount of electric charge. In this exercise, a 600 pF capacitor is charged to 200 V, resulting in an initial charge of \( Q = 1.2 \times 10^{-8} \text{ C} \). Once the charged capacitor is connected to an uncharged capacitor, the charge will distribute between the two.
  • The two capacitors have equal capacitance.
  • The total charge remains constant, conserving charge.
  • Both capacitors will end up with the same voltage across them, known as the common voltage.
This charge redistribution ensures both capacitors share the initial charge equally.
Energy Stored in a Capacitor
The energy a capacitor stores is linked to both its capacitance and the voltage across it. Before any reconnection, the charged capacitor holds energy given by the formula \( E_{\text{initial}} = \frac{1}{2} C V^2 \). For the initial setup:
  • The capacitance \( C \) is 600 pF.
  • The voltage \( V \) is 200 V.
Plugging in these values gives us an initial energy of \( 1.2 \times 10^{-5} \text{ J} \), representing the energy stored due to the electric field in its dielectric material.
Potential Difference
After connecting the charged capacitor to an uncharged one, a new potential difference or 'common voltage' is formed across both capacitors. Since the total charge does not change, the new voltage can be calculated as
  • \( V_{\text{common}} = \frac{Q_{\text{total}}}{C_{\text{total}}} \)
  • The total capacitance \( C_{\text{total}} = 1200 \text{ pF} \).
For this scenario, the common voltage turns out to be 100 V. This essentially means that the potential difference, or voltage, reduces as the initial higher voltage charges both capacitors equally.
Energy Lost in Capacitors
Energy conservation plays a crucial role here. Upon connecting the capacitors, although charges redistribute, some energy seems 'lost'. It's not actually lost but transformed, often into heat due to energy dissipation in the wires or dielectric.
  • Initial energy: \( 1.2 \times 10^{-5} \text{ J} \)
  • Final energy: \( 6 \times 10^{-6} \text{ J} \)
The energy loss is then the difference, calculated as \( E_{\text{lost}} = 6 \times 10^{-6} \text{ J} \). Hence, energy is always conserved, even if it changes form.

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Most popular questions from this chapter

A parallel plate capacitor is charged and then isolated. What is the offect of increasirig the plate separation on charge, potential and capacitance, respectively? a. Constant, decreases, decreases b. Increases, decreases, decreases c. Constant, decreases, increases d. Constant, increases, decreases

A spherical capacitor has an inner sphere of radius \(12 \mathrm{~cm}\) and an outer sphere of radius \(13 \mathrm{~cm}\). The outer sphere is earthed and the inner sphere is given a charge of \(2.5 \mu \mathrm{C}\). The space between the concentric spheres is filled with a liquid of dielectric constant 32. Determine potential of the inner sphere. a. \(400 \mathrm{~V}\) b. \(450 \mathrm{~V}\) c. \(500 \mathrm{~V}\) d. \(300 \mathrm{~V}\)

The plates of a parallel plate capacitor are charged with surface charge densities \(\sigma_{1}\) and \(\sigma_{2}\), respectively. The electric field at points: a. inside the region between the plates will be zero b. above the upper plate and below the lower plate will be zero c. everywhere in the space will be zero d. inside the region between the plates will be uniform and non-zero

A parallel plate air capacitor is charged to \(100 \mathrm{~V}\) and is then connected to an identical capacitor in parallel. The second capacitor has some dielectric medium between its plates. If the common potential is \(20 \mathrm{~V}\), the dielectric constant of the medium is a. \(2.5\) b. 4 c. 5 d. 8

The plates of a parallel plate capacitor have an area of 90 \(\mathrm{cm}^{2}\) cach and are separated by \(2 \mathrm{~mm}\). The capacitor is charged by connecting it to a \(400 \mathrm{~V}\) supply. Then the energy density of the energy stored (in \(\mathrm{Jm}^{-3}\) ) in the capacitor is (Take \(\varepsilon_{0}=8.8 \times 10^{-12} \mathrm{Fm}^{-1}\) ) a. \(0.113\) b. \(0.177\) c. \(0.152\) d. none of these
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