A cylindrical wire of radius 1.0 cm has charge distributed throughout its volume with a uniform charge density of 10 $\mu$ C/m³. The electric field at a point 0.50 cm from the central axis of the wire is

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Quizplus · Accepted Answer

To find the electric field inside a uniformly charged cylindrical wire, we use Gauss's law. The electric field $E$ at a distance $r$ from the axis inside the wire is given by $E = \frac{\rho r}{2\epsilon_0}$, where $\rho$ is the charge density and $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} C^2/N \cdot m^2$). Given $\rho = 10 \times 10^6 C/m^3$ and $r = 0.50 \times 10^{-2} m$, substituting these values gives $E = \frac{(10 \times 10^6)(0.50 \times 10^{-2})}{2(8.85 \times 10^{-12})} = 2.8 \times 10^3 N/C$.

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