Maxwell's electrodynamics is ultimately a theory in $3$ dimensions. Setting

$$\frac{\partial^2 V}{\partial y^2},\frac{\partial^2 V}{\partial z^2}=0$$

by saying that $V$ is independent of $y,z$ doesn't change that. A $1\rm D$ line is still an object in $3\rm D$, and in order to treat it you should use Dirac's delta function. Lets denote by $\lambda\left(x\right)$ the charge per unit length of the wire. Then the volume charge density is given by

$$\rho\left(\vec{r}\right)=\lambda\left(x\right)\delta\left(y\right)\delta\left(z\right)$$

Here the units are

$$\left[\lambda\right]=\frac{\rm C}{\rm m}$$

$$\left[\delta\right]=\frac{1}{\rm m}$$

so you get

$$\left[\rho\right]=\frac{\rm C}{\rm m^{3}}$$

exactly as you've seen. Note that in this case $V=V\left(\vec{r}\right)$, and you can't neglect the coordinates $y,z$.

If you

Maxwell's electrodynamics is ultimately a theory in $3$ dimensions. Setting

$$\frac{\partial^2 V}{\partial y^2},\frac{\partial^2 V}{\partial z^2}=0$$

by saying that $V$ is independent of $y,z$ doesn't change that, and so $\left[\rho\right]=\frac{\rm C}{\rm m^{3}}$ remains correct.

Maxwell's electrodynamics is ultimately a theory in $3$ dimensions. Setting

$$\frac{\partial^2 V}{\partial y^2},\frac{\partial^2 V}{\partial z^2}=0$$

by saying that $V$ is independent of $y,z$ doesn't change that. A $1\rm D$ line is still an object in $3\rm D$, and in order to treat it you should use Dirac's delta function. Lets denote by $\lambda\left(x\right)$ the charge per unit length of the wire. Then the volume charge density is given by

$$\rho\left(x\right)=\lambda\left(x\right)\delta\left(y\right)\delta\left(z\right)$$

Here the units are

$$\left[\lambda\right]=\frac{\rm C}{\rm m}$$

$$\left[\delta\right]=\frac{1}{\rm m}$$

so you get

$$\left[\rho\right]=\frac{\rm C}{\rm m^{3}}$$

exactly as you've seen.

Maxwell's electrodynamics is ultimately a theory in $3$ dimensions. Setting

$$\frac{\partial^2 V}{\partial y^2},\frac{\partial^2 V}{\partial z^2}=0$$

by saying that $V$ is independent of $y,z$ doesn't change that. A $1\rm D$ line is still an object in $3\rm D$, and in order to treat it you should use Dirac's delta function. Lets denote by $\lambda\left(x\right)$ the charge per unit length of the wire. Then the volume charge density is given by

$$\rho\left(\vec{r}\right)=\lambda\left(x\right)\delta\left(y\right)\delta\left(z\right)$$

Here the units are

$$\left[\lambda\right]=\frac{\rm C}{\rm m}$$

$$\left[\delta\right]=\frac{1}{\rm m}$$

so you get

$$\left[\rho\right]=\frac{\rm C}{\rm m^{3}}$$

exactly as you've seen. Note that in this case $V=V\left(\vec{r}\right)$, and you can't neglect the coordinates $y,z$.

If you