Electrostats
ans to -8 hi hai but que me given hai ki electrif field A to B wli line ke along hai to A par potential Zyada hoga to Ve-Vd = Va-Vb ( equipotential surface) ka ans negative kese aaskta hai
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@Gyro Gearloose
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+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Charge of electron is negative.
mm yeah
but for that i hve already taken workdone as negative?
@Real potato it should be +8? just consider the work done on the charge to be q X E.dr.....and equate it to the given value, BUT, E.dr will be replaced by -dv, so the negative of this term and that of q(here -e) will cancel each other to yield positive 8 volt.
dr here, being the distance travelled along line AB (along the E field here too in this case).
Ans is -8 though
just think about it, if your potentiel value is coming negative, it means your potentiel gradient is increasing in the direction A to B, but how is it possible that the E field is then directed in the same direction(that is, up the gradient)!
@Deleted User yehi to mera doubt tha 🥲
.
it will be -8
negative charge of electron has to be considered and along the electric field potential decreases
if your argument states that potentiel decreases along the E field (as it does), the the difference of the initial potentiel to final should be a positive value right?
not -8 as you stated above.....
@Real potato check ur dms i sent a source there as well.
mathematically ans is -8
It is possible that there is an unstated force in the question. Otherwise natural tendency of Electron to move against E-Field renders question incorrect.
@Varun_Arora please look into the question once?
@Deleted User see ur right theoretically, actually my original doubt was this only but mathematically ans is -8 you can check my soln which i uploaded with the que this is exact same solution as given in the ans of the que..
Shouldn't you take work done as positive since displacement and force are in same direction
they are not right?
They are
Along the electric field
electrostatic force on an electron (negatively charged particle) is in the direction opposite to the E field orientation.
Then why is the electron moving along the field
due to some unspecified external force....of which we have not talked here, and we dont need to!
Ok i have to revise electrostatics its getting confusing 💀
electrostatic force is given by F=qE, where E is the electric field at that point, so, suppose our field is in a direction along a unit vector,say r cap, then the negative sign in q, would align the force towards the direction opposite to r cap, which is also opposite to the E field.
then why is your solution wrong
please elaborate?
whats the mistake? can you point it out please?
https://www.toppr.com/ask/question/in-moving-from-a-to-b-along-anelectric-field-line-the-work-done-by-the/
exact same question layout, answer is +4v
It should be +8 only according to me. (Thanos' reason seems correct)
Essence is that, potentiel drop is an attribute of E field distribution in space solely.
Even tho initially i also tripped and thought -8
@Real potato when will this post close? ☠️
Is it sorted ?💀 i hvent figured out yet how is it -8 though
did you ask your teacher @Real potato ? ☠️
this will be enshrined as the doubt that was never closed 🔥 it will go on till eternity ☠️
I asked but i didnt understand a thing he just explained me the solution 🙂↕️
☠️
ye padha? toppr wala?
In moving from A to B along an electric field line, the electric fi...
In moving from A to B along an electric field line, the electric field does 6.4 × 10 19 J of work on an electron. If ϕ1, ϕ2 are equipotential surfaces, then the potential difference V C V A is?A. 4 VB. 4 VC. ZeroD. 64 V
similar thing from byjus
(numbers diff but process exactly same)
similar to what thanos had shared
this is the longest doubt thread (by time) fr ☠️ deserves an entire seperate award 🔥
Yes pdha but isse kuch explanation/ans ni mila is que ka
@Deleted User can you find out any mistake in this soln? This is the exact soln given in the ans of this question
F=-du/dx, so, for writing the work expression f.dx=-du, but the work is already given to be of the certain value in the question, so no need for the extra -ve. in other way you can visualise it like this, the electron has a natural tendency to move opposite to the applied E field, so while travelling opposite to it, its climbing up the potentiel energy hill, and so delta u should have been possible. the better way out,still would be the E relation with potentiel drop, but its directly related with the Force concept so both would work....
I didn’t understand about the tendency of electron to move opposite to E ? Isnt it like force on a negative charge in Electric field is antiparallel to the direction field ? Does it hve to do anything with the negative charge being an electron?
no no i meant that point only, force being anti-parallel with E field direction.
yes i agree it does hve tendency to move opposite but since it is moving along The filed that means the electric field would hve done some negative work na cause the force and displacement are opposite?
yes yes the work done would be negative.
At this point i think there is something wrong in this que theoretically
have you read the toppr answer? it is legit the same question, so are you not convinced yet?
Yes I read actually they took the workdone as +ve
they have not i think, see i think they adjusted the -ve sign in vi-vf instead of vf-vi
yes i think so
but then whats wrong in this soln? They hve also take w = -ve and here in soln i hve also taken -ve 🤔
work done is equal to drop in potentiel energy, so the two negative shouldve cancelled each other.
Ahh i see
we good?
Yes sir
+solved @Deleted User
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