Question

If the linear charge density of a cylinder is 4m14m-1 then electric field intensity at point 3.6 cm from axis is

Answer 1

“4m14m-1 ” what is it ?, I didn't get it. well, I can solve this question by assuming linear charge density of a cylinder is $\lambda$ C/m.

we know,

electric field due to infinitely long straight wire (as cylinderical shape ) at a point r from its axis is given by, $E=\frac{\lambda}{2\pi\epsilon_0r}$

here r = 3.6 cm = 0.036m

so, electric field intensity at the point 3.6 cm from its axis is $E=\frac{\lambda}{2\pi\epsilon_0(0.036)}$

putting $\epsilon_0$ = 8.85 × 10^-12 C²/Nm²

so, E = $\lambda$/(2 × 3.14 × 8.85 × 10^-12 × 0.036)

= 5 × 10¹¹$\lambda$ N/C