IIT JEE , AIEEE Forum - Physics , Electricity

margo

Closest distance of approach of an alpha particle traveling with a certain velocity v to a certain nucleus is x. the distance of closest approach of alpha particle traveling with vel 3v to the same nucleus is>................

a)x/9 b)9x c)x d)x/3

the radius of a hollow metallic sphere is R If pd between its surface and a point at a distance 3R from its centre is V Then electric field intensity at a distance 3R from its centre is.........

a)V/2R b)V/3R c)V/4R d)V/6R

reply early

taruntanuj007

q1) conserve energy the KE is equal to PE at closest approach its pretty easy if u cant do it now nudge me!!!
q2) as per gauss law the sphere can be imagined as a point charge then use coulumb's law again very easy!!! try them or nudge me!!

chinvora

DISTANCE OF CLOSET APP IS r=Ze^2/

m v^2
thus r=x/9

KraniuM

0.5mv^2 =( KZe^2)/r
or

r= (2KZe^2)/mv^2

therfore r is directly proportional to 1/v^2
so x=k/v^2
therefore,
for a speed of 3v, distance is x/9

rate if correct mate!!!

10904him

(a)
Since when the alpha particle as it approaches the nucleus the P.E. increases and K.E. decreases i.e K.E. changes into P.E.
0.5mv^2 = k (q1.q2)/ r = X/r since kq1q2 = X = constant
r=2X/mv^2
When the velocity becomes thrice then the closest distance of approach becomes 1/9 times
Thus ,
x/9 is the answer.

Second Question--

Since V= E.d
Pot on surface = kq/R
Pot at 3R = kq/3R
thus acc to you V=2kq/3R
Thus V at 3R =V/2
Since V= E.d
E = V/6R (d) Ans

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sg_90

the potential difference is between........the surface and a point at a distance 3R frm the centre of the hollow sphere...and field is defined as :-

V/

r

therfore here

V is given as V....and

r=(3R-R)2R

therfore the field intensity at 3R is V/2R

the answer is (A)....

rate me if i am correct!!!!!

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