Question

long straight wire has linear charge density $4\times {10}^{-9}C{m}^{-1}$ .the negative charge q = $-5\mu C$ is at perpendicular distance $\text{4cm}$ from it, The force experienced By the charge is

Answer 1

Step 1: Given that:

Linear charge density($\lambda$)= $4\times {10}^{-9}C{m}^{-1}$

Charge($q$)= $-5\mu C$ =$-5\times {10}^{-6}C$

Perpendicular distance($r$)= $4cm$

Step 2: Calculation of the electric field:

The electric field at a point at a distance r due a straight wire of linear charge density $\lambda$ is given by,

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ ,

Thus,

$E=\frac{4\times {10}^{-9}C{m}^{-1}}{2\pi {\epsilon }_0\times 4\times {10}^{-2}cm}\times \frac{2}{2}$

$E=\frac{8\times {10}^{-9}}{4\times {10}^{-2}}\times \frac{1}{4\pi {\epsilon }_0}$

$E=2\times {10}^{-7}\times 9\times {10}^9$

$(Since,\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9)$

$E=18\times {10}^2$

$E=1800N{C}^{-1}$

Step 3: Calculation of the force on the negative charge:

Force on a charge in an electric field

$F=qE$

Thus,

$F=5\times {10}^{-6}C\times 1800N{C}^{-1}$

$F=9000\times {10}^{-6}$

$F=9\times \times {10}^{-3}N$

(Directed towards the wire.)

Thus,

The force on the charge is $9\times {10}^{-3}N$ .