At what rate must the potential difference between the plates of a parallel-plate capacitor with a2.0μF capacitance be changed to produce a displacement current of1.5A?

Short Answer

Expert verified

The rate of potential is 7.5×105V/s.

Step by step solution

01

Given data

C=2.0μF

id=1.5A

02

Determining the concept

By using the concept of displacement current and by using equation 32-10, write the equation for flux as a function of potential, and from that, find the rate of change of potential difference. The rate of change of potential difference across the plates of the capacitor is displacement current per unit capacitance.

Formulae are as follows:

id=ε0dϕEdt

where, 'i' is current,ϕ is the flux.

03

Determining the rate of change of potential difference

From the equation 32-10,

id=ε0dϕEdt

Where, ϕ=AEtheand theE=Vd.

So,

ϕE=AVd

So,

id=Aε0ddVdt

Where, the capacitance,

C=Aε0d

So, the rate of potential is,

dVdt=idCdVdt=1.52.0×10-6dVdt=7.5×105V/s

Hence, the rate of potential is 7.5×105V/s.

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

The figure shows a circular region of radiusR=3.00cmin which a uniform displacement current id=0.500Aisout of the page.(a)What is the magnitude of the magnetic field due to displacement current at a radial distancer=2.00cm?(b)What is the magnitude of the magnetic field due to displacement current at a radial distancerole="math" localid="1663231983050" r=5.00cm?

Assume that an electron of mass mand charge magnitude emoves in a circular orbit of radius rabout a nucleus. A uniform magnetic fieldis then established perpendicular to the plane ofthe orbit. Assuming also that the radius of the orbit does not change and that the change in the speed of the electron due to fieldis small, find an expression for the change in the orbital magnetic dipole moment of the electron due to the field.

The saturation magnetization Mmax of the ferromagnetic metal nickel is4.70×105A/m . Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is8.90g/cm3, and its molar mass is58.71g/mol .)

A magnetic flux of7.0mWb is directed outward through the flat bottom face of the closed surface shown in Fig. 32-40. Along the flat top face (which has a radius of4.2cm ) there is a 0.40Tmagnetic field directed perpendicular to the face. What are the (a) magnitude and (b) direction (inward or outward) of the magnetic flux through the curved part of the surface?

(a) What is the measured component of the orbital magnetic dipole moment of an electron withml=1?(b) What is the measured component of the orbital magnetic dipole moment of an electron withml=-2?

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