Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 20.0-μF 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-μF 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
Original charge is 0.016 C. Final voltage is 533.33 V. Final energy is 4.2667 J. Decrease in energy is 2.1333 J.

Step by step solution

01

Calculate the Original Charge

The original charge on the 20.0-μF capacitor is calculated using the formula Q=C×V. Here, C=20.0μF and V=800V. Therefore, Q=20.0×106F×800V=0.016C.
02

Calculate the Final Potential Difference

The total charge is conserved, so the charge Q=0.016C is shared between the two capacitors connected in parallel. The equivalent capacitance Ceq is the sum of the two capacitances: Ceq=20.0μF+10.0μF=30.0μF. The final voltage Vf across both capacitors is given by Vf=QCeq=0.016C30.0×106F=533.33V.
03

Calculate the Final Energy of the System

The final energy stored in the system is given by Ef=12CeqVf2. Substitute Ceq=30.0×106F and Vf=533.33V to get Ef=12×30.0×106F×(533.33)2V2=4.2667J.
04

Calculate the Original Energy

The original energy stored in the 20.0-μF capacitor is Ei=12CV2=12×20.0×106F×(800)2V2=6.4J.
05

Calculate the Decrease in Energy

The decrease in energy when the capacitors are connected is the difference between the original energy and the final energy: Eextdecrease=EiEf=6.4J4.2667J=2.1333J.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Electrostatics
Electrostatics is the branch of physics that deals with the study of forces, fields, and potentials arising from static electric charges. In this context, we talk about capacitors, devices used to store electrical energy in an electric field. Capacitors have two conductors separated by an insulator, and when connected to a voltage source, such as a battery, they become charged.

When you consider a charged capacitor, like the initial 20.0-μF capacitor charged to 800 V, it holds a charge calculated by the formula Q=C×V, where C is the capacitance and V the voltage. A charged capacitor has a potential difference across its plates and an electric field inside that exerts forces on any charge placed within the field.

In our problem, this charged capacitor is connected to an uncharged 10.0-μF capacitor. When the connection is made, charge redistribution occurs due to electrostatic forces, causing a new uniform potential difference across both capacitors.
Energy Conservation
Energy conservation is a fundamental principle in physics that states energy cannot be created or destroyed but can only change forms. In electrostatic systems, energy is conserved during charge redistribution. However, you might notice a drop in the system's potential energy when redistribution occurs.

Initially, the 20.0-μF capacitor had energy Ei=12CV2. When connected to another capacitor, the total stored energy appears to decrease. This happens because some energy is transferred to heat and electromagnetic radiation during the rapid redistribution of charge.
The calculated initial energy was 6.4 J. After redistribution, the final energy becomes 4.2667 J. The apparent loss of 2.1333 J exemplifies how some energy escapes as heat due to non-ideal conditions in real capacitors, demonstrating energy conversion rather than loss.
Capacitance Calculation
Understanding capacitance and its calculation is essential when dealing with capacitors. Capacitance C is a measure of a capacitor's ability to store charge per unit voltage, given by C=QV.

In our scenario, two capacitors are connected in parallel after the initial capacitor is charged. The total capacitance Ceq is the sum of the capacitances of both capacitors. So, Ceq=20.0μF+10.0μF=30.0μF.
  • This illustrates how total capacitance in parallel combinations can be found by simple addition.
  • It shows that parallel capacitors share the total charge across them, leading to the same potential difference Vf across each.
The resultant final voltage Vf is calculated using conservation of charge and the formula Vf=QCeq. Thus, the understanding of parallel capacitance is vital for solving such problems accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

A cylindrical capacitor has an inner conductor of radius 2.2 mm and an outer conductor of radius 3.5 mm. The two conductors are separated by vacuum, and the entire capacitor is 2.8 m long. (a) What is the capacitance per unit length? (b) The potential of the inner conductor is 350 mV higher than that of the outer conductor. Find the charge (magnitude and sign) on both conductors.

A cylindrical air capacitor of length 15.0 m stores 3.20 × 109 J of energy when the potential difference between the two conductors is 4.00 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

A parallel-plate capacitor has plates with area 0.0225 m2 separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 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.

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 × 107 V>m. The capacitor is to have a capacitance of 1.25 × 109 F and must be able to withstand a maximum potential difference of 5500 V. What is the minimum area the plates of the capacitor may have?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free