Question

A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a radius of 2.0 m. Find the magnitude of the electric eld at the center of curvature of the arc

Answer 1

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           $\theta = \frac{S}{r}$

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                $\theta = \frac{S}{r}$

                             = $\frac{4}{2}$

                             = $2 radians (\frac{180^{o}}{\pi})$

                             = $114.64^{o}$

Now, we will calculate the charge density as follows.

                 $\lambda = \frac{Q}{L}$

                            = $\frac{18 \times 10^{-9} C}{4 m}$

                            = $4.5 \times 10^{-9} C/m$

Now, at the center of arc we will calculate the electric field as follows.

                 E = $\frac{2k \lambda Sin (\frac{\theta}{2})}{r}$

                     = $\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}$

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.