Question

Electric field outside a long wire carrying charge q is proportional to :

• A. $\frac{1}{r}$
• B. $\frac{1}{r^2}$
• C. $\frac{1}{r^{\frac{1}{3/5}}}$
• D. $\frac{1}{r^{\frac{1}{3/2}}}$

Answer 1

Answer: (A)

$\frac{1}{r}$

Using Gauss Law across the cylindrical Gaussian surface,

$\int E.dS=\frac{q_{enc}}{{ϵ}_0}$
$\therefore E\times 2\pi r\times l=\frac{\lambda \times l}{{ϵ}_0}$
$\therefore E=\frac{\lambda }{2\pi r{ϵ}_0}$

Here $\lambda$ is the linear charge density on the wire.