Electric Field/Magnetic Field Questions

[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 energy 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

 

[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 won't be accelerated, which implies that the kinetic energy of the particel won't 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 $q_1/q_2$.

 

[thepatientmental]

1/ The kinetic energy of a charged particle in a magnetic field is given by the equation KE = 1/2mv^2, where m is the mass of the particle and v is its velocity. In this case, the proton's velocity is perpendicular to the magnetic field, which means that the magnetic force acting on it is always perpendicular to its motion. This results in the work done by the magnetic force on the proton being zero, since the work done is given by the equation W = Fdcosθ, where F is the force, d is the displacement, and θ is the angle between the force and displacement. Since no work is done on the proton, its kinetic energy remains constant.

2/ a/ The electric field at point Z due to charge q1 can be written as E = kq1/d^2, where k is the Coulomb's constant and d is the distance between q1 and Z. This is because the electric field at a point is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge.

b/ The strength of the electric field is equal to zero when the net electric field at that point is zero. This can happen when the electric field due to q1 is equal and opposite to the electric field due to q2. In this case, the ratio of the point charges can be written as q1/q2 = d^2, since the electric field due to q2 is given by E = kq2/(4d)^2 = kq2/16d^2. Setting these two equations equal to each other and solving for q1/q2 gives us the ratio mentioned.

I hope this helps! If you need further clarification or assistance, please let me know. Good luck with your problems!