Question

A -1.2 C charge moves along a 3 N/C electric field. How much work is done on the charge if it moves a distance of 0.4 meters?

Answer 1

The work of the electric field is equal to

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

Given that

$\phi_2>\phi_1$

Finally write

$A=q \cdot(\phi_1-\phi_2)=q \cdot(-\Delta\phi_)=-q \cdot(E \cdot \Delta S)=-(-1.2 \cdot 3 \cdot 0.4)=1.44 J$