Why Does a Rigid Electric Dipole Move in an Electric Field?

[MaddenDude]

First:
A rigid electric dipole is free to move in the electric field in the pic...
http://img53.imageshack.us/img53/6950/untitled8kz.jpg
Which one of the following phrases most accurately describes the initial motion of the dipole if it is released from rest in the position shown?
The Answer is: "It moves to the left"

Why does it move to the left?


Second:
A metal sphere is radius 8 cm is charged to a potential of -500V. With what velocity must an electron be fired toward the sphere if it is to just barely reach the sphere when started from a position of 15 cm from the center of the sphere?



Help would be greatly appreciated. These were test questions, and I got them wrong.

 

[nrqed]

First:
A rigid electric dipole is free to move in the electric field in the pic...
http://img53.imageshack.us/img53/6950/untitled8kz.jpg
Which one of the following phrases most accurately describes the initial motion of the dipole if it is released from rest in the position shown?
The Answer is: "It moves to the left"

Why does it move to the left?

The electric force on the left charge is in what direction? The electric force on the right charge is in what direction?
Now, at which of the two points is the electric field the strongest? (hint: the density of the E field lines tell you something about the magnitude of the E field at a point).

 

[MaddenDude]

thank you!

 

[nrqed]

Second:
A metal sphere is radius 8 cm is charged to a potential of -500V. With what velocity must an electron be fired toward the sphere if it is to just barely reach the sphere when started from a position of 15 cm from the center of the sphere?



Help would be greatly appreciated. These were test questions, and I got them wrong.

${1 \over 2} m v_i^2 - e V_i = {1 \over 2} m v_f^2 - e V_f = -e \times -500 Volts$

Outisde of the sphere, the electric potential varies with distance the same way as the electric potential produced by a point charge, which is $k_e q / r$. The key point is that it varies inversely with the distance. Since it's -500 V at 8 cm, you can easily find the potential at 15 cm. And you're done.


Patrick