Question

If the total charge on a thin rod of length $0.4$is$2.5\mathrm{nC}$, what is the magnitude of the electric field at a location$1\mathrm{Cm}$from the midpoint of the rod, perpendicular to the rod?

Short answer

$1.12\times {10}^4\mathrm{N}/\mathrm{C}$

Step-by-step solution

Identification of the given data

The given data is listed below as-

• The length of the thin rod is, $L=0.4\mathrm{m}$
• The charge of the thin rod is,
• The location of the electric field is,

Significance of the electric field

The electric field is a region in which a charged particle is able to exert force on another charged particle.

The concept of the electric field gives the magnitude of the electric field.

Determination of the magnitude of the electric field

The equation of the magnitude of the electric field is expressed as,

$E=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{Q}{\sqrt[X]{X^2+{\left({\displaystyle \frac{L}{2}}\right)}^2}}\right)$

Here,$\frac{1}{4{\mathrm{\pi \epsilon }}_0}$is the constant of electric field, $Q$is the charge of the thin rod, $X$is the location of the electric field and$L$ is the length of the thin rod.

For $\frac{1}{4{\mathrm{\pi \epsilon }}_0}=8.99\times {10}^9\mathrm{N}.\mathrm{M}/{\mathrm{C}}^2$, $Q=2.5\mathrm{nC}$, $X=1\mathrm{cm}$and $L=0.4\mathrm{m}$.

$E=8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{c}}^2\times \left(\frac{25\mathrm{nC}}{\sqrt[1\mathrm{cm}]{{\left(1\mathrm{cm}\right)}^2+{\left({\displaystyle \frac{0.4\mathrm{m}}{2}}\right)}^2}}\right)$

$=8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \left(\frac{2.5\mathrm{nC}\times {\displaystyle \frac{1\times {10}^{-9}\mathrm{C}}{1\mathrm{nC}}}}{1\mathrm{cm}\times {\displaystyle \frac{1\mathrm{m}}{100\mathrm{cm}}}\sqrt{{\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)}^2+{\left({\displaystyle \frac{0.4\mathrm{m}}{2}}\right)}^2}}\right)$