Question

An electron is released from rest at$x=2m$ in the potential shown in Figure Q26.6. Does it move? If so, to the left or to the right? Explain.

Short answer

The electron makes a Rightwardmovement.

Step-by-step solution

Given information and formula used    

Given :

Initial position of electron :$x=2m$

Graph -

Theory used :

Electric fieldis the negative slope (“derivative” or “gradient”) of the potential.

$\overrightarrow{E}=-\frac{\Delta V}{\Delta x}$

Potential energy is related to electric potential by the charge. It only makes sense if you know the sign & magnitude of the test charge you want to use.

Determining the direction of the electron 

The electron makes a Rightwardmovement.

Because the electric field is the inverse of the electric potential's slope, the electric field will be nonzero and pointing to the left at $x=2m$.

As a result, the electron will experience a net force to the right and accelerate in that direction.