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A charged particle of charge Q is held fixed and another particle of mass m and charge q is released from distance r. Find The impulse of the force exerted by the external agent on the fixed charge by the time distance between Q and q becomes 2r.

Maira Kapoor , 7 Years ago

Grade 12

2 Answers

Rohan Kadkol

Last Activity: 7 Years ago

$I=\int Fdt$

$I=\int \frac{kQq}{r^{2}}dt$

$I=\int \frac{kQq}{r^{2}}dt . \frac{dr}{dr}$

$I=kQq\int \frac{dr}{r^{2}v}$

where v is velocity as dr/dt=v

$E_{initial}=E_{final}$

$0+\frac{kQq}{R}=\frac{1}{2}mv^{2}+\frac{kQq}{r}$

$v=(\frac{2kQq(\frac{1}{R}-\frac{1}{r})}{m})^{1/2}$

Substituting and simplifying the integral

$\sqrt{}\frac{mkQqR}{2}\int_{R}^{2R}\frac{dr}{r^{\frac{3}{2}}\sqrt{r-R}}$

r-R = t^2

dr=2t dt

$\sqrt{}\frac{mkQqR}{2}\int_{R}^{2R}\frac{2tdt}{t(t^{2}+R)^{\frac{3}{2}}}$

t=root(R)tanx

dt=root(R)sec^2(x)dx

$2\sqrt{}\frac{mkQqR}{2}\int_{R}^{2R}\frac{\sqrt{R}\sec ^{2}x}{R^{\frac{3}{2}}\sec^{3}x}$

$\frac{2}{R}\sqrt{}\frac{mkQqR}{2}\int_{R}^{2R}\frac{dx}{secx}$

$\frac{2}{R}\sqrt{}\frac{mkQqR}{2}\int_{R}^{2R}cosx$

$\sqrt{}\frac{2mkQq}{R}sinx$

Now put values of x as t and t as r and then do the usual definate integral method of F(2R)-F(R)

Vikas TU

Last Activity: 7 Years ago

Impulse(J) =Integral( Fdt) ..(I cannot put the fundamental sign pls get it)

= integral(kq q/r dt) = kq q vital( dt/r ) = kq q fundamental( dt/r (dr/dr) ) ( mulytiply and separate by dr)

= kq q necessary( dr/r v) (since v = dr/dt ) .....(1)

in any case, we can't coordinate eq 1 as v is a variable and we need to express it as an element of r before continuing with the

coordination...

This can be accomplished by moderating vitality..

Ei = Ef

kq q/R + 0 = mv/2 + kq q/r

or, on the other hand v = [ (2kq q/m)(1/R - 1/r) ] ....( call this condition 2 )

substitute eq 2 in 1 to get

J = Integral ( dr/{ r [ (2kq q/m)(1/R - 1/r) ] } ).....(3)

presently you can coordinate it from R to 2R.

The joining is bit lengthy..but not tidious thus I am maintaining a strategic distance from it here ....but I can give u a clue...

Subsequent to simpifying eq 3 utilize the substitution (r-R)=t

after that substitute t = Rtan(X)...the fundamental will lessen to the shape .... An Integral(cos (X) dx )...(A is a constant).which can without much of a stretch assessed utilizing the development of cos(3x).

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