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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.

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.

Grade:12

2 Answers

Rohan Kadkol
22 Points
6 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
14149 Points
6 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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