What Causes the Negative Sign in Electrostatic Potential Energy Calculations?

[latentcorpse]

another pretty simple question that i can't see:

a 1d harmonic oscillator of mass m carries an electric charge q. a weak uniform, static electric field of magnitude E is applied in the x direction. what is the classical electrostatic potential energy for a point particle at position x.

ans: W=-qEx

i said $W=\frac{1}{4 \pi \epsilon_0} \frac{q q_p}{x}$ where $q_p$ is the point particle.

now $E=\frac{1}{4 \pi \epsilon_0} \frac{q_p}{x^2} \Rightarrow W=qEx$

i have two questions,
(i) why am i missing a negative sign?
(ii)was i correct to use $q_p$ is my expression for E? surely if $q_p$ is located at x and i have taken E to be the electric field at x then there's a problem there, unless both charges were just meant to be taken as q?

 

[gabbagabbahey]

(i) why am i missing a negative sign?

$\frac{1}{4 \pi \epsilon_0} \frac{q q_p}{x}$ represents the work done by the fields in bringing the charges in from infinity. The energy stored in the fields therefore decreases by that amount (conservation of energy) from its initial value (Usually defined to be zero when the charges are infinitely far away from each other).

(ii)was i correct to use $q_p$ is my expression for E? surely if $q_p$ is located at x and i have taken E to be the electric field at x then there's a problem there, unless both charges were just meant to be taken as q?

If you are calculating the field due to $q_p$, then yes, it will appear in your answer. However, $x$ is your source point, so you want to use a different variable to represent your field point (You need only worry about field points on the x-axis), the electric field falls off proportional to the square of the distance between your field point and source point.