-1
$\begingroup$

Say you place a positively charged particle between two plates with an uniform electric field between them. The particle will accelerate towards the negatively charged plate with a constant acceleration, and the work done on the particle by the field is defined by:

$$\Delta W = qE \Delta x$$

Where $\Delta W$ is the work done, $q$ is the particle's charge, $E$ is the uniform field strength and $\Delta x$ is the distance the particle has moved.

From my understanding, $\Delta W$ should be equivalent to the particle's loss of absolute electrical potential energy and it's gain in kinetic energy. However, as an answer to a textbook question with this setup, $\Delta W$ was defined as the gain in EPE and also the gain in KE.

However, isn't that impossible, because if the particle gains $\Delta W$ in KE and in EPE then it will gain $2\Delta W $ in overall mechanical energy, which is impossible because only $\Delta W$ work has been done?

Is there an error in my textbook, or am I making a conceptual error somewhere?

Improve this question
$\endgroup$
3
  • $\begingroup$ The particle loses EPE and gains KE. It does not gain both. $\endgroup$
    – sammy gerbil
    Commented Jul 26, 2017 at 12:23
  • $\begingroup$ That's what I thought. So then the textbook has an error. $\endgroup$
    – Pancake_Senpai
    Commented Jul 26, 2017 at 13:02
  • 1
    $\begingroup$ If the textbook says that both EPE and KE increase in this situation then Yes the textbook has an error. $\endgroup$
    – sammy gerbil
    Commented Jul 26, 2017 at 13:06

1 Answer 1

Reset to default
0
$\begingroup$

In electrostatics, the work done on the charge in moving it from position $A$ to position $B$ is

enter image description here

Here, we choose the mechanical force $Fm$ such that it balances the electric force $qE$ at each point along the path. What does this say about the kinetic energy?

Improve this answer
$\endgroup$
1

Not the answer you're looking for? Browse other questions tagged or ask your own question.