Question

A force of 0.037 N is required to move a charge of 35.9 µC a distance of 24 cm in an electric field. What is the size of the potential difference between the two points?

Answer 1

We write an expression for the work on the movement of a charge in an electric field

$A=q(\phi_2-\phi_1)=q \cdot\Delta\phi$

The same work can be expressed as

$A=F\cdot S$

then

$F\cdot S=q \cdot\Delta\phi$

from here

$\Delta\phi=\frac{F\cdot S}{q}=\frac{0.037\cdot 0.24}{35.9 \cdot 10^{-6}}=247.354 B$