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In electrostatics, electric potential is a scalar quantity that describes the amount of potential energy per unit charge at a particular point in space. It is often denoted by the symbol \( V \) and is measured in volts (V). The electric potential due to a point charge or a distribution of charges gives an idea of how much work is needed to move a charge within an electric field. In the given problem, the potential function is \( V(x) = 3x - 2x^2 - x^3 \), a polynomial function of degree three.

To find where the potential equals zero on the x-axis, you set \( V(x) = 0 \). Solving the equation \( 3x - 2x^2 - x^3 = 0 \) involves determining the roots of this cubic equation, which tells us at which points the potential vanishes. By factoring, we find \( x(x^2 - 2x + 3) = 0 \), giving \( x = 0 \) as the only real solution. This means the electric potential is zero at \( x = 0 \) on the x-axis.

The electric field is a vector quantity that represents the force experienced by a unit positive charge placed in a particular location. It has both magnitude and direction, and is denoted by \( \mathbf{E} \). In our problem, the electric field is derived from the electric potential function. The relationship between them can be expressed as \( \mathbf{E} = -abla V \), where \( abla V \) is the gradient of the potential.

Since the problem is one-dimensional, the electric field \( E(x) \) is simply the negative derivative of the potential with respect to \( x \). Differentiating \( V(x) = 3x - 2x^2 - x^3 \), we get \( \frac{dV}{dx} = 3 - 4x - 3x^2 \). Hence, the electric field \( E(x) = - (3 - 4x - 3x^2) = -3 + 4x + 3x^2 \). This expression allows you to calculate the electric field at any point on the x-axis.

Solving cubic equations is a key mathematical skill in many physics problems, including those involving electric potential. A cubic equation generally takes the form \( ax^3 + bx^2 + cx + d = 0 \). To solve the equation \( x^3 - 2x^2 + 3x = 0 \) in our scenario, we first factor out an \( x \), leading to \( x(x^2 - 2x + 3) = 0 \).

This gives one solution immediately: \( x = 0 \). The quadratic \( x^2 - 2x + 3 \) remains, but solving it using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) shows it has no real roots, since \( b^2 - 4ac = (-2)^2 - 4(1)(3) = 4 - 12 = -8 \), which is negative. Thus, there are no real solutions from the quadratic part; therefore, \( x = 0 \) is the only real root of the equation, revealing the points on the x-axis where the cubic function equals zero.