39Disclaimer : I am only an amateur in physics.
We have to find the acceleration of the block by force equations.
It is the horizontal direction wrt the block we are concerned with.
qEcos30° + mgsin30° = macos30°
=> √3qE2 + mg2 = √3ma2
=> √3qE + mg = √3ma
=> a = qEm + g√3
Now we have the height h from which the block begins from rest, and the acceleration of the block.
h = ut + 0.5at²
=> h = 0 + (qE2m + g2√3) * t²
Looks like my answer's coming wrong..you didn't specify which side the positive plate is on and which side the negative one is..
1+ve plate is on the left side...question is from RSM
39
Pritish Chakraborty
·2010-05-26 07:56:02
lol ab khud karo....tabhi qE + mg aa raha hai aur minus nahi :P
1
ajoy abcd
·2010-05-26 08:02:28
Thanks...got it(and sorry for incomplete information)
39
Pritish Chakraborty
·2010-05-26 08:07:19
What about the second one? We somehow have to use V0 but I'm not getting..
1u know na
E = - dVdr
usng this will help i guess
39
Pritish Chakraborty
·2010-05-26 08:11:50
I did that but not getting...got an integration and I'm getting something like : V = -Exx - Eyy...we have to use V0 too isn't it?
39
Pritish Chakraborty
·2010-05-26 08:15:58
Yes I'm not getting the V0. Evidently some mistake I've made.
1
ajoy abcd
·2010-05-26 08:34:24
could this be the process????.... .
somebody please confirm...
39
Pritish Chakraborty
·2010-05-26 11:04:50
Yes this is it! We had to take potential difference..lol.
The upper limit is (x, y, 0) btw...but it's alright.
And the lower limit is (0,0,0).