11Tension =Eq
Q1. Multiple answer correct:
There exist an electric field in space given by the relation \vec{E} = E_{0}x^n\hat{i}.
Consider an imaginary cube of side l, centred at (\frac{l}{2},\frac{l}{2},\frac{l}{2}) whose faces are parallel to xy, yz and zx planes. Select the correct alternative(s).
(a) If n=0, then charge enclosed by the cube is zero
(b) If n=1 then the cube encloses some positive charge
(c) If n= -1 then the cube encloses some negative charge
(d) If n=2 then the charge enclosed by the cube is zero.
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10 Answers
106Q2. A block of mass m and charge q is attached to a rod of same mass m at one end. The other end of the rod is hinged at O. The system lies on a smooth horizontal table. An electric field E is switched on in perpendicular direction to rod and in the plane of the table.
Calculate the tension developed at the mid point of the rod, when the rod becomes parallel to electric field.
11See C means E= E0/x
rite it means that the electric field decreases from 0 to l along the x-axis
Only negative charge inside the cube can make this possible. First place the cube on the left side of the centre . At 0 the electric field will increase and at l it will decrease that is wat is wanted. Now place it on the right hand side of the centre. At 0 the electric field increases and at l it decreases.
106virang try using gauss's law. the electric field at x=0 is not defined. So, while calculating will u take LHL or RHL? Yes the electric field decreases from 0+ to ∞ but at 0- the field is -∞ so u can say it increases also.
My answer is that u cant say because the electric field itself is undefined. at x=0
Ya ur answer is correct according to solutions
but i have doubt on this point only.
My answer is that u cant say because the electric field itself is undefined. at x=0
3gud to be bak.......how r u asish .....wassup............
had gone outta station for a month and half
the second question ...................[methodology]
see u have a torque abt COM = Eql/4cos@ [@=angle betw the rod and vertical ]
I(com)WdW/d@ = Eql/4 * cos@ .....
cross multiply integrate frm 0 to pi/2 ................ get W................
now we can calculate tension by segmenting the rod and integratin the centrifugal force ........ to that add Eq and the centrifugal force of the mass m..........
9asish u r confusing urself
ofc E cant be defined at x =0
eg even in well known case of a point charge u analyse and see