no yaar, its not that. what actually is happening is that we have taken that whole cartesian is an infinite plane. now obviously charge density would remain same. then, apply formula sigma/2(epsilon). now actually we dont have 4th quadrant so, net is by 3only. by symmetry contribution is equal. therefore, 3/4.
@sambhav.. E as used by me in that post is simply any electric field existing in that region.. but actually it doesn't matter.. what matters is the Z-component of the field E..
HI someone please help me in this question, i have been trying this for over 2 hours.
A PARTICLE MOVES WITH A CONSTANT SPEED 1.5 m/s ALONG A PATH GIVEN BY x=y^2 - ln(y). GIVE THE ACCELERATION OF THE PARTICLE IN VECTOR FORM AT THE POSITION Y=3m.
answer given is -.225i +4.5j (or i,j cud be reveresed, its just given in component form in the answer)
@ blog admin answer is 3/4(epsilon) i verified this analogy by assuming plane to be 3 quarters of charged ring and then integrating for a disc. Result was same.
@ delhietie
bro this answer must be wrong i tried a lil hook n crook but the vector sum of velocities deduced by this acceleration would never sum up to 1.5 and acceleration in x direction must not be negative, instead in y direction must be negative as the particle bends towards x axis rapidly.
i got this answer a = 2.111 i - 0.372 j
my steps were 1) differentiate twice and put value 3 2) this acceleration will be directed in direction of motion at that instant. 3) Take components acos@ and a sin@ 4) find out tan@ = slope on tangent at that point by differentiating one time. 5) Find sin@ and cos@ from tan@ 6) find acos@ + asin@
But i cant assure you of its correctness took me a day but could not cross examine it.
@delhiite, I didn't solve your question but I know how to do it. I will give you the method. diff given equation wrt time. dx/dt=vx dy/dt=vy vx^2+vy^2=v^2 i.e constant.
now find vx in terms of v and x vy in terms of v and y. Now again diff to get equation in ax. I am just telling you the way. You have to solve yourself, you will get the answer
3(epsilon)/4
ReplyDeletei have seen this earlier
but im myself not convinced how we can treat infinite plate like this
@sambhav: i guess u mean 3/(4(epsilon0)
ReplyDeleteas for your doubt.. for calculating the work here, we need only the z component of the electric field.. thus done!
how are we applying infinite plate formula?
ReplyDeletewasn't that formula meant for a point symmetrically placed close to somewhat the center of a really, really large plate?
yahan to vo point corner par placed hai?
no yaar, its not that. what actually is happening is that we have taken that whole cartesian is an infinite plane. now obviously charge density would remain same. then, apply formula sigma/2(epsilon). now actually we dont have 4th quadrant so, net is by 3only. by symmetry contribution is equal. therefore, 3/4.
ReplyDeleteI hope it is clear now!
amit u r incomplete..
ReplyDeletesee guys obviously the z component of field contributed by d 3 quadrants will be the same.. direction as well as magnitude..
the difference in direction will be there only for the x-y plane component of the field..
now for the work.. ie, qV we integrate E.dr.. but dr=dz k cap.. hence we need only the z component of the field!
yaar plz help with this one..
ReplyDeletehttp://anujkalia.blogspot.com/2010/12/i-love-this-one.html
shivanker, what's E in your solution
ReplyDeleteand i myself am waiting for anuj bhaiya on that
shivanker, acc. to your soln you will take E each of quadrant and net you will get 3E. from where will you get 4 in denominator.
ReplyDeleteI have replied to love this one post
every quadrant will contribute sigma/(8.epsilon0) in +z direction..
ReplyDelete@sambhav.. E as used by me in that post is simply any electric field existing in that region.. but actually it doesn't matter.. what matters is the Z-component of the field E..
ReplyDeleteshivanker main bhi to yahi keh raha hun ki every field will contribute sigma/8(epsilon)
ReplyDeleteanswer is -(3/4(epsilon))
HI
ReplyDeletesomeone please help me in this question, i have been trying this for over 2 hours.
A PARTICLE MOVES WITH A CONSTANT SPEED 1.5 m/s ALONG A PATH GIVEN BY x=y^2 - ln(y). GIVE THE ACCELERATION OF THE PARTICLE IN VECTOR FORM AT THE POSITION Y=3m.
answer given is -.225i +4.5j (or i,j cud be reveresed, its just given in component form in the answer)
thaks in advance.....!!!
This comment has been removed by the author.
ReplyDelete@ blog admin
ReplyDeleteanswer is 3/4(epsilon)
i verified this analogy by assuming plane to be 3 quarters of charged ring and then integrating for a disc.
Result was same.
@ delhietie
bro this answer must be wrong
i tried a lil hook n crook but the vector sum of velocities deduced by this acceleration would never sum up to 1.5
and
acceleration in x direction must not be negative, instead in y direction must be negative as the particle bends towards x axis rapidly.
i got this answer a = 2.111 i - 0.372 j
my steps were
1) differentiate twice and put value 3
2) this acceleration will be directed in direction of motion at that instant.
3) Take components acos@ and a sin@
4) find out tan@ = slope on tangent at that point by differentiating one time.
5) Find sin@ and cos@ from tan@
6) find acos@ + asin@
But i cant assure you of its correctness
took me a day but could not cross examine it.
where did u find this question ??
hi Shivam.t aka shiveminem
ReplyDeletethanks a lot man for giving your time!!
(yaar, that was a dash in front of .225, not a minus)
see i think your step 2] is wrong(it is given velocity is constant)
i am mainly having a problem in bringing the answer into vector form.
some one please help
@delhiite, I didn't solve your question but I know how to do it. I will give you the method.
ReplyDeletediff given equation wrt time.
dx/dt=vx
dy/dt=vy
vx^2+vy^2=v^2 i.e constant.
now find vx in terms of v and x
vy in terms of v and y. Now again diff to get equation in ax.
I am just telling you the way. You have to solve yourself, you will get the answer
thanks amit, got it....
ReplyDeleteThe answer to the question posted by me is sigma/4(epsilon0);
ReplyDeleteHint: remember how you calculated the radial component of the field due to a semi-infinite wire.
The solution to the question asked by delhite is as follows:
1)Calculate the radius of curvature of the path at y=3. (I assume you know the formula).
2)Get the direction cosines of the normal i.e (-dx/dy).
3)Accn. is in the direction of the normal vector and is v^2/r in magnitude.
@anuj bhaiya:
ReplyDeletebhaiya usi method se kiya tha..
vertical component of the field will be 3/4*sigma/2(epsilon0)
so answer is 3sigma/4(epsilon0)
m i going wrong somewhere?
ReplyDeleteThis comment has been removed by the author.
ReplyDelete