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

ΔV = 7.5 V

Step-by-step 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=(25\cdot 10^(-6))/(50 \cdot 10^(-6))=0.5 (V)/(m)`
We can now calculate the potential difference:

`\Delta V=E\Delta L=0.5 (V)/(m)\cdot15m=7.5V`