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## Homework Statement

Suppose a particle of mass

*m*and charge

*q*is moving with velocity

**v**(t) to the right in the second quadrant of a coordinate system C and whose position is described by the vector

**r**(t) pointing from the origin to the particle's location in C. A charge with magnitude

*Q*is fixed at the origin of this coordinate system. What would be the path of the moving charge as it is deflected by the fixed, stationary charge

*Q?*

## Homework Equations

$$\vec F_e=\frac {qQ}{4\pi\epsilon_0[r(t)]^2} \hat r$$ $$\vec F =m\vec a $$

$$\hat r = \frac {\vec r}{r(t)}$$

$$\vec r = x(t) \hat i + y(t) \hat j$$

$$\vec a = \frac {d^2x}{dt^2} \hat i + \frac {d^2y}{dt^2} \hat j$$ $$r(t) = \sqrt{x^2+y^2}$$

## The Attempt at a Solution

The electric force exherted on

*q*by the fixed charge will be equal to Sir Isaac Newton's Second Law of Motion (since

*Q*will impart an acceleration to the components of

*q).*Thus, $$\frac {qQ}{4\pi\epsilon_0[r(t)]^2} \hat r = m\vec a $$ and after some substituting, $$\frac {qQ}{4\pi\epsilon_0[r(t)]^3} [x(t) \hat i + y(t) \hat j] - m[\frac {d^2x}{dt^2} \hat i + \frac {d^2y}{dt^2} \hat j]= 0$$ Let $$ k=\frac {qQ}{4\pi\epsilon_0 m} $$ and then, to satisfy the preceeding equation, it must be that $$\frac {d^2x}{dt^2}=\frac {kx}{\sqrt{(x^2+y^2)^3}} $$ $$\frac {d^2y}{dt^2}=\frac {ky}{\sqrt{(x^2+y^2)^3}} $$ My thinking is that, if the parametrized coordinates

*x*and

*y*can be solved for, then we can trace out the path given by the vector

**r**. If we let $$x=sin t$$ and $$y=cos t$$ the system of equations is solved, but this yields a circle of radius one for all values of

*t.*Is my approach correct? Does this system of differential equations have an analytical solution? Should I be using also the Lorentz Force Law since the moving particle will emit radiation? I would appreciate any input on any information anybody may have and put me on the right track if I am making any mistakes or have any misconceptions. Thanks! :)