# Electric Field/Magnetic Field Questions

1. Jun 22, 2005

### lightuplightup

I urgently need help with these two problems:

1/ A proton enters a magnetic field in an evacuated chamber.Its initial velocity is at right angles to the magnetic field. Explain why the kinetic enrgy of the proton stays constant.

2/ There are 2 point charges separated by a distance of 4d in a vacuum. Point Z is situated on a line between q1 and q2 at a distance 'd' from q1.

a/ Write an expression in terms of q1 and d for the electric field at point Z due to charge q1.

b/ the strength of the electric field is equal to zero.
What is the ratio of the point charges in terms of q1:q2

Any help/guidance is greatly appreciated- especially for the second question. Thanks a lot

2. Jun 22, 2005

### Vegeta

1). You know that the force from a magnetic field acting on a charge with a velocity, is given by
$$\vec{F}=q(\vec{v}\times\vec{B})$$
Hence the velocity is perpendicular to the force, and there the particel wont be accelerated, which implies that the kinetic energy of the particel wont change, only the direction og the velocity vector will change.

2) Consider this in 1-dimension, the electric field from point charge $q_1$ is
$$\vec{E}_1=\frac{1}{4\pi\varepsilon_0}\frac{q_1}{r^2}\hat{r}$$
The unit vector is $\hat{r}=\frac{\vec{r}}{|\vec{r}|}$ where $\vec{r}$ can be expressed as $\vec{r}=x\vec{i}$, because we only consider 1-dimension, therefor you get, $|\vec{r}|=\sqrt{x^2}=|x|$, $\hat{r}=\frac{x}{|x|}\vec{i}$,
$$\vec{E}_1=\frac{q_1}{4\pi\varepsilon_0}\frac{x}{|x|^3}\vec{i}$$
You can calculate the E-field from the other charge in a similar way.
Is assume that you mean the E-field equals 0, in points Z, right? Then solve the equation $\vec{E}_1+\vec{E}_2=0$ for $q_1/q_2$.