- #1
Kyle Nemeth
- 25
- 2
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! :)