# Charged box and spring in uniform electric field--when does the box stop moving?

I'm having a lot of trouble trying to figure this out: a box of mass m and positive charge Q is attached to a spring of spring constant k, which is in the equilibrium position. They sit horizontally on a table, with the box to the right of the spring. The other end of the spring is attached to a wall. The whole setup is immersed in a uniform electric field E, which points rightward, away from the spring, so that when the box moves the spring is stretched.

I'm not sure exactly what happens in this situation. I thought the box stopped moving when the spring force and electric force were equal, but then I realized that of course the box still has inertia and will move along at a constant speed. But then the spring will continue to cause it to accelerate... right? Does that mean it oscillates back and forth at a new point of equilibrium?

Also, how can you find the maximum amount the spring is stretched? It seems like you'd have to use calculus, but this problem is from an algebra-based textbook. (I only know a tiny amount of calculus).

The only way I can think of to solve this problem is setting the spring potential energy at x equal to the electric potential energy... but since the field is uniform, the electric potential energy is infinite, right?

(If it's any use to anyone, this is problem 9, Chapter 16, page 593 in Serway Vuille College Physics, Tenth Edition.)

• The potential energy is not infinite. What is the expression for the work done by an uniform electric field? – nasu Mar 26 '17 at 20:16

Returning to your setup, we can now see that the box would oscillate about some point of equilibrium. To find it, you just need to write the force equation. By definition of electric field, we have $$F_{elec.}=Q\cdot E$$ Choosing the positive $x$ direction for the electric force and letting the spring equilibrium be at $x_{0}=0$ for simplicity, we get:
$$F_{spring}+F_{elec.}=-kx+Q\cdot E=0$$ therefore $$x_{eq.}=\frac{Q\cdot E}{k}$$ This means that if the system should come to rest, it will stop exactly at $x_{eq.}$, because otherwise the forces would not cancel and by Newton's 2nd law there will be some accelaration and thus motion.
As for the maximum stretch, we can see that about $x_{eq.}$ the net force is simply the difference between the spring and the field, where to the right of $x_{eq.}$ it is in the negative $x$ direction (spring wins) and to the left it is to the positive $x$ direction (field wins), always proprtional to the distance from $x_{eq.}$. Therefore, we have nothing more than a simple harmonic oscillator, and this means that the maximum stretch is exactly on the other side of $x_{eq.}$, i.e. $$x_{max}=2\cdot x_{eq.}$$
• This is because the oscillation is always about the equilibrium point. By turning on an electric field, you "recalibrate" your system such that $x_{eq.}$ is what shown in the answer. And relative to it - you indeed start your motion in the "opposite maximum". – Yoni Mar 26 '17 at 22:29
It is just like a vertical spring-mass system with the constant force on the mass due to a uniform gravitational field equal to $mg$ except in your problem the constant force $QE$ is produced by a uniform electric field in the horizontal plane. In both cases the other force acting on the mass is due to the spring.