A cylindrical wire of radius 1.0 cm has charge distributed throughout its volume with a uniform charge density of 10.0 $\mu$ C/m³. The electric field at a point 0.50 cm from the surface of the wire is

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To find the electric field at a point outside a uniformly charged cylindrical wire, we use Gauss's law. The electric field $E$ at a distance $r$ from the axis of the cylinder (where $r$ is greater than the radius $R$ of the cylinder) is given by $E = \frac{\rho R^2}{2\epsilon_0 r}$, where $\rho$ is the charge density, $R$ is the radius of the cylinder, and $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} C^2/N \cdot m^2$). Given $\rho = 10.0 C/m^3$, $R = 1.0 cm = 0.01 m$, and $r = 1.5 cm = 0.015 m$ (since it's 0.5 cm from the surface of the wire), substituting these values gives an electric field of approximately $3.8 \times 10^3 N/C$.

A Cylindrical Wire of Radius 1 μ\muμ C/m3 ×\times× 103 N/C ×\times× 103 N/C

A Cylindrical Wire of Radius 1 $\mu$ C/m³ $\times$ 10³ N/C $\times$ 10³ N/C