There are 2 uniformly charged spheres (call them charge clouds):
Sphere 1: volume charge density: P, radius: 1, center: (0,0,0).
Sphere 2: volume charge density: -P, radius: 1.5, center: (1,1,1).
Find the electric field in their region of intersection.
Answers welcome.
come on people, or is it really 'too easy'?
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ReplyDeletei tried 'superposition' and got
ReplyDelete"E= P1*D1/ 3e + P2D2/3e "
where, P= modulus of volume charge density
D= dist of respective centres from the plane of intersection
e= epsilon........
pls tell if its right( m not too sure)
sambhav it is easy
ReplyDeletespecially if uve done a similar questn b4
d only trick i can think of is to do it with vectors
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ReplyDeletethe answer is
ReplyDeleterho*c1*c2/3*eps?
correction-- , not C1*C2..but C1C2(vector)
ReplyDeleteanonymous is correct
ReplyDelete@shaan, had P1 and P2 been their 'actual' densities, you'd have been correct
E= P1*D1/ 3e + P2D2/3e would have become P/3e(D1-D2), which becomes P(C1C2)/3e
D1 and D2 are position vectors of the any point in the region of intersection from the centers of the spheres, and by vector triangle, D1-D2 is the vector C1C2
and what if i say, there is just a single sphere of given charge density, with another spherical 'cavity' completely inside it?
ReplyDeletealso, do these apply to gravitation as well?
i think this is a new 'too easy' record.
ReplyDeletecud u draw the electric field lines here
ReplyDeletethanks sambhav, i missed that..
ReplyDeletefor single sphere , it wud remain same..
ie E=P(C1C2)/3e
also, this shud apply to gravitation, by the perfect analogy..
whatsay??
yeah. only the constants vud change for gravitational force
ReplyDeletethe first case wouldn't, we don't actually have poles in gravitation, although the second would.
ReplyDeleteobviously sambhav
ReplyDeleteyes, obviously :P
ReplyDelete