Electric field and point charge problem

[zinedine_88]

Homework Statement

A particle with mass m and chare, q<0, moves in the field of a POINT charge Q>0, which is fixed in the origine of a cartesian coordinate system. The particle starts at REST in a distance R on the x-axis of a cartesian coordinate system : r(o)(vector) = Rix(vector), V(o) = 0

a) ... what is the force on the particle with charge "q" . Write down the equation of motion.

b) Show that the particle moves in a straight line along the x - axis!

C) prove the energy conservcation LAW :

[(m/2)(dx/dt)^2] + qQ / 4(pi)[E(not)]abs(x) = E = const!

d) how long does it take for the particle to reach the center ( where the charge Q sits)

Homework Equations

a) - i thing that the force is F = Qq/4(pi)(E(not)(R^2)times(-ix) unit vector , because it is moving from the right to the center... since Q is fixed! but tell me if i am right or wrong please..

b) I have no idea how to prove that mathematically...obviously there is no reason that q should move along the Y axis... but don't know what to do...

c) it says that i have to use the time derivative of the expression above and to use the equation of motion for x to show that it vanishes.. i don't get that HINT :( we have to assume that X is bigger than zero

d) says that i have to use the energy conservation law from part c...

there is nothing like that in our book...idk..

please help me

 

[Mindscrape]

b) Hello, you have a good start. You already showed that part b was true when you calculated the force. The force vector only has an x component, so the particle must only move along the x axis.

c) Remember that $F = m \frac{d^2 x}{dt^2}$. What did you get for part c as it is?

d) Yes, if you use energy conservation then you can figure out when the particles collide.

 

[Moetasim]

a) The force on a particle with charge q in an electric field E is given by F = qE. In this case, the electric field at a distance R from a point charge Q is E = Q/4(pi)(E(not)(R^2)) times the unit vector in the direction of the electric field. Therefore, the force on the particle is F = -Qq/4(pi)(E(not)(R^2)) times the unit vector in the opposite direction of the electric field.

The equation of motion for a particle with mass m is given by F = ma, where a is the acceleration. Substituting the force from above, we get ma = -Qq/4(pi)(E(not)(R^2)) times the unit vector in the opposite direction of the electric field. Since the particle is moving along the x-axis, the acceleration in the y and z directions is 0, so we can write the equation of motion as ma = -Qq/4(pi)(E(not)(R^2)) times the unit vector in the x direction.

b) To show that the particle moves in a straight line along the x-axis, we need to show that the acceleration in the x-direction is constant. From the equation of motion above, we can see that the acceleration is constant because it depends only on the fixed values of Q, q, and E(not). Therefore, the particle moves in a straight line along the x-axis.

c) To prove the energy conservation law, we need to use the time derivative of the expression given in the problem: [(m/2)(dx/dt)^2] + qQ / 4(pi)(E(not))abs(x) = E = const. Taking the time derivative, we get [(m/2)(d^2x/dt^2)](dx/dt) + qQ / 4(pi)(E(not))dx/dt = 0. Using the equation of motion from part a, we can substitute for (d^2x/dt^2) and (dx/dt), giving us [(m/2)(-Qq/4(pi)(E(not)(R^2)))(-Qq/4(pi)(E(not)(R^2))) times the unit vector in the x direction] + qQ / 4(pi)(E(not))(Q/4(pi)(E(not)(R^2)) times the unit vector in the