Question

A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential difference? Recall that Fe = qE.

Answer 1

Answer:

ΔV = 7.5 V

Explanation:

Answer 2

Electric potential differnce is defined as:
$\Delta V=\int_{a}^{b}E\cdot dl$
If we asume the electric field is constant we get:
$\Delta V=\int_{a}^{b}E\cdot dl=E\Delta L$
We need to calculate the electic field strenght:
$F=qE$
$E= F/q\\ E=\frac{25\cdot 10^{-6}}{50 \cdot 10^{-6}}=0.5 \frac{V}{m}$
We can now calculate the potential difference:
$\Delta V=E\Delta L=0.5 \frac{V}{m}\cdot15m=7.5V$