Point charges q and --q are located at the
vertices of a square with diagonals 2l as shown
. Find the magnitude of the electric
field strength at a point located symmetrically
with respect to the vertices of the square at a
distance x from its centre.
12)A thin nonconducting ring of radius R has a linear charge
density λ=λ0cos\theta, where λ0 is a constant, θ is the azimuthal
angle. Find the magnitude of the electric field strength on the axis of the ring as a function of the distance x from its
centre.
1in the first ques i cant understand "electric
field strength at a point located symmetrically
with respect to the vertices of the square at a
distance x from its centre"
131) The point will be at "x" height frm the centre, which makes it a pyramidal case...
1ohh achaa ab samajh aya thnxx............neone for second one..i tried but not gettin the desired answer....
12nd ques
azimuthal angle refers to angle θ from any diameter . visualise it as for trigenometric angles 0 to 2π
the circular ring has two symmetrical right and left semicircular halves but right half has +ve sign and left one has -ve .
find the field at the point x distance above centre of ring and double it.
11
E1 = 2 E cos theta = 2 k q l[x2 + l2] 3/2 along AC
E2 = 2E cos theta = 2 k q l[x2 + l2] 3/2 along DB
Therefore, net field E0 = √E1 2 + E 2 2 = q l √2pie epsilon (x2 + l2) 3/2
1thnxx bahiiya i was having prob in understanding the situation thnxx for posting the solution...................for second can t get it.is it like this anirudh....
1yeah sanchit it' s like that only the charge distribution is not same throughout it varies by the relation given.
13)Identical thin rods of length 2a carry equal charges,
+Q, uniformly distributed along their lengths. The
rods lie along the x axis with their centers separated by
a distance of b> 2a. Show that the magnitude
of the force exerted by the left rod on the right
one is given by
F=\left(\frac{ke Q^2}{4a^2} \right)ln\left(\frac{b^2}{b^2-4a^2} \right)
here ke means ke......
1This shudn't be tuff. Consider a +ve test charge in the centre. Find individual field and add them vectorially. Or find components. Where exactly are you facing problem?
1@avik
its not double integration (volume under surface)
it just integration carried out two times
in the first integration treat x as constant
in second treat y as constant
1hmm got it.......thnxxx to all ...ll post more doubts soon............ abhi doubts baki haii mere dost :P
49
now if the charge is in the space b/w the 2 charges then the forces on individual charges gets added...
if it is in space other than the mid region then also the force cannot be zero
since force could have been zero in either of the 2 sides of the axial line...if the charges were unequal...
so a and b cancelled..
torque can be zero if θ=0°
else torque s finite...
1pXE can be zero, hence C)
Note: the orientation of the torque is not mentioned.
1achaa u mean if theta =0 then torque is zero other ways it is a finite value......hmm aa gaya samajh mein thnxx shubho nd ray.........[1][1][1][1]
