Differences between Induced Electric Fields and Electrostatic Fields | EMI | Physics | Khan Academy

Let's explore all the differences between Induced Electric fields and Electrostatic fields!
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Created by Vibhor Pandey

[@Wanderlust_91]

I never understood this in myall attempts of neet UG exam but watching this 12 min video has made my concepts crystal clear... thank u so so much guruji...u truly made my day 😀

[@howtuber3754]

Excellent explanation

[@kotaprodigy]

The best video on induced E field. Summary: 1) induced electric field is NON CONSEVATIVE due to +ve work done every time the charge completes loop. 2) E.dl = -dv = - (-dΦ/dt) 3) Induced E field is tangential to direction of induced current.

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[@aryanraj441]

It was helpful

[@ronycb7168]

Thank you clear explanation

[@vikasjangir514]

Great sir❤

[@Wanderlust_91]

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[@vordark304007]

Faraday Law(Non Conservative Fields) vs Kirchoff Voltage Law (Conservative Fields)

[@chandachowdhury5180]

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[@muhamadadem4430]

Lagrangian mechanics and Computational physics lecture video

[@piyushjayswal7734]

At b I think that potential should be 15 joule as delta v defined as -f.dl

[@griffithfimeto3387]

Particle ar point a in the direction of electric field has 10 J/c then moving against electric field to point b has 5J/C , how ? Isn't the oppsite true ?

[@sitaramar13]

A particle is moving in direction of force. Will potential energy increase or decrease

[@suleymanaghamoglanli4439]

decrease

[@sitaramar13]

@@suleymanaghamoglanli4439 right. Nice.

[@ronycb7168]

Sir shouldnt mechanical energy be greater at point b because of the of direction of field lines

[@nirbhayupadhyay8492]

Yes I also think so

[@debadityanath5558]

sir is non uniform electrostatic field conservative as well?

[@Ettoyeaz]

Yes, every electrostatic electric field created by a distribution of charges is conservative in nature. This is because the electrostatic electric field is defined as the Coulomb's force exerted on a test charge Q divided by that same charge Q and the following statements: 1)The Coulomb's force (k*q1*q2/r^2) that is exerted by a charge on another charge is conservative. 2)The total force on a charge Q exerted by a distribution of charges is equal to the sum of every single force that every charge in the distribution exerts on Q. Then, because of statement 2), when you calculate the path integral on a closed loop of the electrostatic field, you can write it as the sum of every single integral of every electric field caused by every charge in the distribution. Because of 1), every single electric field is conservative, which implies that its path integral on a closed circuit is zero. This means that the total integral must be zero, which implies that the electrostatic electric field caused by a distribution of charges is always conservative.

[@Ettoyeaz]

To answer your question more specifically, the example of the uniform field given by an infinite plane of charges is just an easy case to understand the idea, but no matter the geometry of the distribution or how complicated the field lines are, if the electric field is the electrostatic one defined by Coulomb's law, it is always conservative.

[@NavSci]

Im not from khan academy But yes , if its not produced by a changing magnetic field then it is conservative because take an example of point charge it is non uniform , any path you take for a charge particle the work done is positive (if to move against the field's) is negative(if moving with the field's force) which at the end will cancel each other out