2.7. The electric field intensity at each point in space is directed towards the origin, and its magnitude is proportional to the distance of the point from the origin. Determine the change in energy of a charged particle when it is moved from $A(0,0,0)$ to $B(1,1,1)$. The charge of the particle is $Q$. The proportionality constant of the field is taken to be 1.
Solution. According to the problem, in each point of space $R_{3}$, a vector $\dot{\mathcal{A}}=(P, Q, R)=(-x,-y,-z)$ is defined.

The change in the energy of the particle is equal to the work of the field during the movement from $A$ to $B$ with the opposite sign. We will calculate the work of the field using a line integral of the second kind:

$$
\begin{aligned}
& \Delta E=-q \int_{A B} P d x+Q d y+R d z=q \int_{A B} x d x+y d y+z d z= \\
& =\frac{q}{2} \int_{A B} d\left(x^{2}+y^{2}+z^{2}\right)=\left.\frac{q}{2}\left(x^{2}+y^{2}+z^{2}\right)\right|_{(0,0,0)} ^{(1,1,1)}=\frac{3 q}{2}
\end{aligned}
$$

This field has a potential $U=\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)$, so the change in energy does not depend on the type of curve connecting points $A$ and $B$.

Problems on determining a function from its total differential.