The electrostatic force $$(\vec{F_1})$$ and magnetic force $$(\vec{F_2})$$ acting on a charge $$q$$ moving with velocity $$v$$ can be written :

This question asks for the correct expressions for the electrostatic force and magnetic force acting on a charge $$q$$ moving with velocity $$\vec{v}$$.

Recall that the electrostatic force on a charge $$q$$ in an electric field $$\vec{E}$$ is given by:

$$\vec{F_1} = q\vec{E}$$

This force acts in the direction of $$\vec{E}$$ for a positive charge and opposite to $$\vec{E}$$ for a negative charge. Importantly, this force does not depend on the velocity of the charge.

The magnetic force on a charge $$q$$ moving with velocity $$\vec{v}$$ in a magnetic field $$\vec{B}$$ is given by:

$$\vec{F_2} = q(\vec{v} \times \vec{B})$$

Key features of this force:

- It involves the cross product of velocity and magnetic field, not a dot product.

- The order matters: it is $$\vec{v} \times \vec{B}$$, not $$\vec{B} \times \vec{v}$$ (they differ by a negative sign).

- This force is always perpendicular to both $$\vec{v}$$ and $$\vec{B}$$, so it does no work on the charge.

- If the charge is stationary ($$\vec{v} = 0$$), the magnetic force is zero.

Comparing with the given options shows that option (1) gives $$\vec{F_1} = q\vec{E}$$ and $$\vec{F_2} = q(\vec{V} \times \vec{B})$$, which matches both formulas correctly.

Option (2) incorrectly uses $$\vec{B}$$ for the electrostatic force.

Option (3) has the cross product in the wrong order ($$\vec{B} \times \vec{V}$$).

Option (4) incorrectly uses dot products instead of cross products.

The correct answer is Option (1): $$\vec{F_1} = q\vec{E}$$, $$\vec{F_2} = q(\vec{V} \times \vec{B})$$.