The potential at a point x (μ m) due to some charges situated on...

The electric potential is given by $V(x) = x^2 + 1.5 x + 2$

The electric field is related to the potential by the equation $E = -\dfrac{dV}{dx}$

Differentiating the given potential with respect to $x$:

$E = -\dfrac{d}{dx}(x^2 + 1.5 x + 2) = -(2x + 1.5)$

Substituting $x = 3\ \mu\text{m}$:

$E = -(2(3) + 1.5) = -7.5\ \text{V}/\mu\text{m}$

The magnitude of the electric field is $7.5\ \text{V}/\mu\text{m}$ and the negative sign indicates that it is directed in the negative X-direction.

Answer: $7.5\ \text{V}/\mu\text{m}$ and in the -ve X - direction.