Class 12th - Potential Difference - Relation with Electric Field | Tutorials Point

Potential Difference - Relation with Electric Field
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Lecture By: Mr. Pradeep Kshetrapal, Tutorials Point India Private Limited

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Watch Playlist on Electrostatic Potential and Capacitance:
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[@Kratosfrost]

For all who do not know, the master of his subjects...had been long gone to rest in eternity peacefully ❤️

[@venuvs100]

What??

[@djaqt2]

Nooooo 😢

[@LostUfo669]

Bro u can express urself in ur mother tongue, Sir is Indian... & please work on ur english.

[@dhivyabharath5962]

i thought my english was poor until i read the replies of this comment

[@JamaahmedMohamud]

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[@simplechora812]

Sir is no more..
.we miss you sir😢😢😢😢

[@simplechora812]

@Charan tej M yes

[@MahaLakshmi-ii1hc]

thanks a lot for these kind of detailed lectures..Your videos are helpful to so many students who self prepare for neet examination inorder to under the concepts first

[@ERYX_YT]

Mone ni polichu ketto.....

[@user-pj1wv1ns9x]

Ahda

[@abhijitburagohain4447]

Thanks a lot sir ❤️.... love from ASSAM 💕

[@emd70999]

Hoi nki

[@tirthapaul7979]

There is a common mistake which is haunting me again and again ( though Sir H.C. Verma has cleared that mathematically) At first it was mentioned charge is taken from B to A. To do that one must apply an equal external force from B to A against the electrostatic force. So Direction of applied force and direction of movement is same i.e. form B to A. So the angle theta must be zero. So students please ask these question.

[@priyankathakare5708]

The force in consideration is not the external force but the force of the electric field. So, the angle is 180 degrees as done by CBSE.

[@MahaLakshmi-ii1hc]

Inorder tp displace the charge from point B to point A we need to apply force which is more than the force exertd by electric field and in the direction opposite to that of force exerted by electric field...then only we can displace it..if you apply equal as that force exerted by electric field and oposite to it , then the charge will be stationary...so if we apply force on the charge in a direction so that it makes some angle with the electric field line,the component of the force which is applied by us in the direction along the electric field line will be more than the force exerted by the electric field on the charge at point B

[@DenisWainaina-du3ou]

this content can last forever since physics does not change

[@sanchitpriyadarshi6882]

Sir, if we will carried a unit charge from point B to A in uniform electricfield lines then the force will remain same at both of points A and B because F=qE and Electric field strength at every point in uniform electricfield lines is same. Sir my questions is why have you considered small distance which is tends to zero while force is same and we apply work= Force*displacement because force is uniform.

[@koulikmanna2948]

Electric field is always a function of distance (r²). Higher the distance lower the electric field . And force is q x E. So force will not be uniform also.

[@taladiv3415]

I came here looking for an explanation about the electric field as the gradient of the potential.
Got out with the nice pointer about the minus sign in another formula 8:30 .

[@tubercambo7590]

Thank teacher from cambodia🇰🇭

[@user-pj1wv1ns9x]

Damn

[@dipun4849]

Sir,I have a doubt
A......B
Va-Vb=-E.dr
=>Vb>Va
But actually Va>Vb.

[@anshulpatel3695]

Old always be gold♥️☺️

[@surendrakverma555]

Excellent lecture Sir. Thanks and Regards

[@vanithadevadoss4943]

Each and every concept is cleared!

[@firdosefathima3778]

Thank you sir it is helpful

[@loveisback3462]

Love ur teachng sir it is slw and understanding....

[@sivakamasundaris3152]

Awesome explanation sir, thank you

[@Imran52Feb]

If we move a unit positive charge from A to B the work done will be negative since the work is being done by the field's force and not by any external force hence dW = -dV= - (Va-Vb) =E dr and hence E = - dV/dr . Am I correct ? PLEASE COMMENT YOUR VIEW, SIR.

[@johnwi-l_l-iamsf3763]

He is dead man ....

[@johnwi-l_l-iamsf3763]

For what u ask , i think , when we allow a charge to go with the field the potential difference will be taken as Vb-Va , so it u will get a negative value , according to what i think , what u mention as " work been negative when taken by field " is a theory , theory which is obtained from this .......
Since charge is taken from A to B , the potential difference is Vb-Va .... just what i think kk ....

[@divyadinani7703]

Thank you sir . These videos are quite helpful,Easily understood.

[@dreamer-cj8bu]

Thanks sir

[@shashwat2037]

watch the at 1.5 times and save ur time

[@lijaraniswain7987]

Thank you so much sir

[@laiqasaleem7045]

The concept is explained in a very simple and understandable manner,sir. Thank you so much for these lectures. But I have doubt. If 'qe' is the the force experienced by the charge and if we want to move it from B to A, then we have to apply a force which is greater than 'qe' ,right?If it's equal to 'qe',then the charge wont move,right?

[@hi-kj1ih]

Excellent sir

[@KAjyoti]

Clear cut explanation

[@vikalp305]

a true lagend for me... ❤

[@rizzurizz1608]

Thanks a lot sir 💯

[@mdsabirhussaineee1158]

Sir Mai apka ye electrostatic ka 34 video dekha aur samhaj Mai bhi agya kya ye 34 video board exam ke liye sufficient hai

[@mdsabirhussaineee1158]

Please sir reply jarur dena

[@mathstricksforcompetition5888]

@@mdsabirhussaineee1158 he is very busy person, he wont .i m sure

[@MohdAly-fo6eu]

So was it enough for boards ??

[@sridurka9846]

Thank u very much sir...sir will u explain these concepts with solving some problems related to this..

[@sofialyric]

Why it is dE=-dv/dr????? E is not small why?? Plzz explain

[@trevorcroteau1253]

@@sofialyric It is not dE but rather E=-dV/dr. Potential (V) results from the fact that E is conservative and it’s curl is 0. Therefore E is the divergence of a vector, the potential. It’s simply an intrinsic property and is always true for E in electrostatics.

[@sofialyric]

@@trevorcroteau1253 ok I understand lil bit but when we have to calculate potential when E is non uniform then why do we not take dE to integrate like
dv=$E.dr ,, why not dv=$dE.r or something
Here above why the magnitude of E (which is non uniform) we take it E instead of dE whyy????? Because E is non uniform we should use small factor of E to integrate !!!!?? But no????? Why??? Plzzz tell I am confused with this thing and also ii noticed that in gausses law if we have to calculate electric flux by integration because of E ( bcoz it's non uniform ) we again write it as - d@= $E.dA plzzz tell me 😭😭😭😭😭

[@trevorcroteau1253]

@@sofialyric Okay lots of questions. First it can be written to V = -$E.dx. It is equivalent, and works like the formula W=$F.dx from mechanics courses. As far as the electric flux, this is Gauss’ law. It’s only useful in high degrees of symmetry where E is uniform over the surface (planar, cylindrical, spherical). In these cases it is useful because we can pull E out of the equation and we are left with $dA which is of course just area. In other cases, it is best to use some other trick.

[@sofialyric]

@@trevorcroteau1253 ohh thanks for your suggestion for gauss law !! It is used for symmetrical surface so that we can pull out E from integration and makes it easier to do!!! And as you said that for mechanical course the formula used to calculate the work when force is variable that is W=$F.dx ok that's correct but I can't understand that if (like you said that we use this formula f.dx to calculate work ) force is variable then we should integrate by chunks of F but we write Complete F inside the integration rather than df !!!!! This is the problem I am facing to understand !!!!!

[@kovendhanko3794]

Super sir.. kindly reply me... what is different between E & vector E

[@siddarajpatil629]

Both are electric field

[@arpitapatil9298]

Really thanks sir

[@farjadraza4446]

Best explanation ever heard for such a relation

[@nirupamam2814]

Very nice..

[@rudranshmishra8869]

Aap bhut acha samjhate hai sir aap ko very very thank you

[@faysalkhan3685]

I just love the lesson, sir

[@shubhammeena5023]

Why it is +200 as formula is -E×dr

[@jenifarpeter2061]

Me too have that same doubt

[@MahaLakshmi-ii1hc]

anyone plz clear my doubt,if the charge displaces by means of our applie dforce then the workdone by us is force applied by us *displacement...then why sir is multiplying force exerted by electric field and the displacement provided by us inorder to find the workdone on charge...plz clear my doubt

[@saraswatidevi2406]

best teacher of India.

[@sumansahoo9998]

Sir F=-qE because against the electric field charge moving

[@sajneebeauty3610]

work done is against the opposite direction bt sir is tlking abt the force which is in the direction of E.F that is y its +ve and not -Ve.and if m wrong then plz do correct me

[@mamoonayasmeen4392]

Thanks sir . From Pakistan

[@dae1925]

😢 RIP Sir

[@johnwi-l_l-iamsf3763]

What happened??

[@prabhakarrao116]

Sir is dead

[@sumansahoo9998]

Thanku sir

[@aounalimirza484]

sir g you are very cute sir and teach very good love u sir gggg

[@LMinPrep]

Flawless

[@UmeshKumar-gr1dh]

Namskar sir

[@melodyshow2008]

What about in non uniform electric field????

[@PrakashKumar-kz9hm]

Sir please provide me notes...

[@chrisgeorge5609]

RIP sir

[@ganapathibhat5360]

What u r talking

[@prabhakarrao116]

Sir is dead

[@suryavg8204]

Guru nav kelo questionge ans madu

[@anupampandey7254]

Sir

[@AjaYSaiNi-ul6no]

Sir please provide note

[@anjuyadav4045]

bsc classes physics chemistry and mathematics

[@emd70999]

Thank u sir

[@anupampandey7254]

Thanks very much

[@AjayMishra-js5si]

Thanks 🙏 sir

[@arunalamichhane1217]

Thank u so much sir