The potential energy function (in J) of a particle in a region of space is given as $$U = (2x^2 + 3y^3 + 2z)$$. Here $$x, y$$ and $$z$$ are in meter. The magnitude of $$x$$-component of force (in N) acting on the particle at point $$P(1, 2, 3)$$ m is:

We need to find the magnitude of the x-component of force at point P(1, 2, 3) given the potential energy function $$U = 2x^2 + 3y^3 + 2z$$.

The force is the negative gradient of the potential energy:

$$\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$$

Therefore, the x-component of force is:

$$F_x = -\frac{\partial U}{\partial x}$$

$$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x$$

(The terms $$3y^3$$ and $$2z$$ are treated as constants when differentiating with respect to $$x$$.)

$$F_x = -4x$$

At $$x = 1$$:

$$F_x = -4(1) = -4 \text{ N}$$

The magnitude of the x-component of force is $$|F_x| = 4$$ N.

The correct answer is Option (3): 4 N.