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

q=lωsinωt Was derived by using the relationship l=dp>dtbetween the current and the charge on the capacitor. In Fig. 31.9a the positive counterclockwise current increases the charge on the capacitor. When the charge on the left plate is positive but decreasing in time, isl=dp>dtstill correct or should it bel=dp>dt? Is l=dp>dt still correct when the right-hand plate has positive charge that is increasing or decreasing in magnitude? Explain.

Short Answer

Expert verified

When the charge on left plate is positive but decreasing in time so time rate of charges dqdtwill be negative value and current direction will be clockwise but current equation will be same l=dqdt.

Step by step solution

01

Define convention of current equation

The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery.

02

Explanation


The current is positive when charge on left plate is increasing satisfyingl=dq>dt.


When charge on left plate is positive decreasing with time so l=-dq>dtbut since current is flowing in anticlockwise direction so l=dqdtwill be positive.

When the charge on right plate is positive which is increasing or decreasing with time, the current equation l=dqdtwill be same via same reasoning of left plate.

Thus, the current equation will always be same for any plate.

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!

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

Question: A positive point charge is placed near a very large conducting plane. A professor of physics asserted that the field caused by this configuration is the same as would be obtained by removing the plane and placing a negative point charge of equal magnitude in the mirror image position behind the initial position of the plane. Is this correct? Why or why not?

A particle of mass 0.195 g carries a charge of-2.50×10-8C. The particle is given an initial horizontal velocity that is due north and has magnitude4.00×104m/s. What are the magnitude and direction of the minimum magnetic field that will keepthe particle moving in the earth’s gravitational field in the samehorizontal, northward direction?

In Europe the standard voltage in homes is 220 V instead of the 120 used in the United States. Therefore a “100-W” European bulb would be intended for use with a 220-V potential difference (see Problem 25.36). (a) If you bring a “100-W” European bulb home to the United States, what should be its U.S. power rating? (b) How much current will the 100-W European bulb draw in normal use in the United States?

A 1.50- μF capacitor is charging through a 12.0-Ω resistor using a 10.0-V battery. What will be the current when the capacitor has acquired14of its maximum charge? Will it be14of the maximum current?

When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

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