Question

The electric field $5.0\mathrm{cm}$from a very long charged wire is ( $2000\mathrm{N}/\mathrm{C}$, toward the wire). What is the charge (in $\mathrm{nC}$) on a $1.0$-cm-long segment of the wire$?$

Short answer

The charge on a segment of the wire is$-0.0556\mathrm{nC}$.

Step-by-step solution

Formula for linear charge density

Electric field,

$\mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Linear charge density,

$\mathrm{\lambda }=\frac{\mathrm{q}}{\mathrm{l}}$

Given segment $∆\mathrm{l}$, the linear charge density will be:

$\mathrm{\lambda }=\frac{\mathrm{q}}{\mathrm{\Delta l}}$

Derivation for charge

Electric field,

$\mathrm{E}=\frac{\left(\frac{\mathrm{q}}{\mathrm{\Delta l}}\right)}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

$\mathrm{E}=\frac{\mathrm{q}}{\left(\mathrm{\Delta l}\right)2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

Rearrange the value,

$\mathrm{q}=\left(\mathrm{\Delta l}\right)2{\mathrm{\pi \epsilon }}_0\mathrm{r}·\mathrm{E}$

Calculation of of the wire

Charge is,

$\mathrm{q}=(0.01\mathrm{m})2\mathrm{\pi }\left(8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}·{\mathrm{m}}^2}\right)(0.05\mathrm{m})\left(2000\frac{\mathrm{N}}{\mathrm{C}}\right)$

$=5.56\times {10}^{-11}\mathrm{C}$

$\mathrm{q}=\left(5.56\times {10}^{-11}\mathrm{C}\right)·\left(\frac{1.0\times {10}^9\mathrm{nC}}{1\mathrm{C}}\right)=0.0556\mathrm{nC}$

Because the electric field is oriented toward the wire, the charge on the wire will have a negative charge.

$\mathrm{q}=-0.0556\mathrm{nC}$