11can u expain why the path ll b lik this
THIS QUESTION WAS SOLVED BY MY SIR LAST YEAR ,BUT STILL I AM NOT SATISFIED WITH THE SOLUTION
I HOPE I WILL GET SOME AWESOME ANSWERS IN THIS FORUM
a particle of mass m , charge q is released from -∞ along +ve x axis
with a velocity v
another charge Q is fixed at (0,-d)
find the deviation angle of the particle after a long time
P.S v is very high
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UP 0 DOWN 0 2 9
9 Answers
1Its trjectory turns into a straight line when the stationary charge is collinear to it.
So it will be some what like this
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11Alternately without applying Gauss Law, we can also evaluate the force componenets along X-axis and Y-axis.
cos \theta = \frac{x}{\sqrt{x^2+(y+d)^2}}
And sin \theta = \frac{y+d}{\sqrt{x^2+(y+d)^2}}
Applying \frac{Fsin\theta }{m}= \frac{v_ydv}{dy} & integrating we have
\int_{0}^{y}{\frac{K'q^2(y+d)dy}{\left\{ x^2+(y+d)^2\right\}\sqrt{x^2+(y+d)^2}}}=\int_{0}^{v}{vdv}
Needless to mention, in this integral x is taken as a const. For simplifying this we can have the substitution \left\{(y+d)^2+x^2 \right\}=m
So that gives \boxed{ v_y= \sqrt{\frac{K''}{\sqrt{d^2+x^2}}-\frac{K'''}{\sqrt{(y+d)^2+x^2}}}}
K', K'' , k''' are all constants that can be calculated.
Finally we also have on similar basis \boxed{ v_x= \sqrt{v^2-\frac{2K''''q^2}{\sqrt{x^2+(y+d)^2}}} }
(Minus sign neglected as it denotes the decrease).
tan\alpha = \frac{v_y}{v_x}
Now on solving the equations \frac{\partial tan\alpha }{\partial x}=0 & \frac{\partial tan\alpha }{\partial y}=0, we get the reqd (x,y)....from which tan\alpha can be calculated, which is the reqd angle.
1reviving the thread !!
but what kind of trajectory will trace
i think branch of hyperbola due to following reasons
1.energy of system is +ve
2. the differential equation will be similar to planetary motion
instead of working in rectangular co ordinate ,
we can also work in polar coardinats and take advantage of force being entirely radial
the diferential equations are :
\frac{\partial^2r }{\partial t^2}-r\left(\frac{\partial\theta }{\partial t} \right)^2 =\frac{Qq}{4\pi\varepsilon_0}\frac{1}{r^2}............\circled{1} \\ \\ \text{angular momentum conservation } \\\\ r^2\frac{\partial \theta}{\partial t}=v_0d ........2 \\ \\ \text{someone please eliminate t from 1 and 2 , i am not good in diff equations :P}
1continuing .............
\text{after solving the differnetial equation we finally arrive at the equation}\\ \\\frac{1}{r\left(\theta \right)}=\frac{-Qq}{4\pi \varepsilon _0mv_0^2d^2}+A\cos(\theta -\theta _0)\\ \\\\\text{using initial conditions and the fact}\ v_0 \rightarrow \infty, \\\\ \text{we get}\\ \\0=\cos(\theta _0)\rightarrow \theta _0=\frac{\pi}{2}\\ \sin\theta =\frac{Qq}{4\pi\varepsilon _0mv_0^2d^2A}\\\\ \text{Since v }\rightarrow \infty,\theta \text{must tend to 0, so we can take } \\\\ \ \sin\theta \approx \tan\theta \approx \theta \\ \\ \text{trying to find value of A ,but no other condition is given}
