A block of charge q and mass m is connected to a spring of constant k. An electric field E exists parallel to the ground. The block is released from rest from a unstreched spring. Find Maximum displacement?

Apr 29, 2018

$\frac{2 E q}{m}$

Explanation:

Newton's 2nd law:

• $F = m a$

Here:

• $F = E q - k x$

The spring linearly opposes displacement from the equilibrium position, hence the negative term and the harmonic oscillation.

Hence, equation of motion:

$E q - k x = m \ddot{x}$

• $\implies \ddot{x} + \frac{k}{m} x = \frac{E q}{m}$

General solution:

• $x = A \cos \omega t + B \sin \omega t + \frac{E q}{k}$, where $q \quad {\omega}^{2} = \frac{k}{m}$

With IV's:

• $x \left(0\right) = 0$

$\implies x = B \sin \omega t + \frac{E q}{k} \left(1 - \cos \omega t\right)$

• $x ' \left(0\right) = 0$

$x ' = \omega B \cos \omega t + \omega \frac{E q}{k} \sin \omega t \implies B = 0$

So the governing equation is:

• $x = \frac{E q}{k} \left(1 - \cos \sqrt{\frac{k}{m}} t\right)$

Because $- 1 \le \cos \theta \le 1$:

• $0 < x < \frac{2 E q}{k}$