Question

The electric field at any point $P(x>b)$ outside the outer cylinder is :

• A. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$ radially inward
• B. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$ radially outward
• C. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$ co-axially
• D. none of these

Answer 1

Answer: (B)

$\frac{\lambda }{2\pi {\epsilon }_{0}r}$ radially outward

Using guass law we can say that electric field at any point will be because of total charge inside the cylinder which is $\lambda l$
$\oint E\cdot ds=\frac{q_{in}}{{\epsilon }_0}$;
total charge $q_{in}=\lambda l$
$E\left(2\pi rl\right)=\frac{\lambda l}{2\pi {\epsilon }_{0}rl}$
or $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
and electric field due to inside charge will be in radially outward direction.