A circular rod has a radius of curvature $\mathbf{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{9}\mathbf{.00}\mathbf{\text{cm}}$and a uniformly distributed positive charge$\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.25}\mathbf{\text{pC}}$and subtends an angle$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.40}\mathbf{\text{\hspace{0.17em}rad}}$.What is the magnitude of the electric field that$\mathbf{\text{Q}}$produces at the center of curvature?

1. Radius of curvature of a circular rod,$\mathrm{R}=9.00\text{\hspace{0.17em}cm}$
2. Uniformly distributed positive charge,$\mathrm{Q}=6.25\text{\hspace{0.17em}pC}$
3. The subtended angle,$\mathrm{\theta }=2.40\text{\hspace{0.17em}rad.}$

Using the formula of the electric field of a charged rod, we can get the value of the electric field that is using the given data.

Formulae:

The magnitude of the electric field of a charged circular rod,$\mathrm{E}=\frac{\mathrm{\lambda sin}(\mathrm{\theta }/2)}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}$ (i)

Where, R = the distance of field point from the charge

q = charge of the particle

θ = the angle subtended that is expressed in radians

The length of the circular arc,$\mathrm{L}=\mathrm{r\theta }$ (ii)

The linear charge density of the body, $\mathrm{\lambda }=\left|\mathrm{Q}\right|/\mathrm{L}$ (iii)

“Electric field of a charged circular rod” illustrates the simplest approach to circular arc field problems. Thus, the value of the electric field is given using equations (ii) and (iii) in equation (i) as follows:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{arc}}=\frac{(\left|\mathrm{Q}\right|/\mathrm{R\theta })\text{sin}(\mathrm{\theta }/2)}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{R}}\\ =\frac{\left(\left|\mathrm{Q}\right|\right)\text{sin}(\mathrm{\theta }/2)}{2{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{\theta }}\\ =\frac{2\left(6.25\mathrm{x}{10}^{-12}\text{\hspace{0.17em}C}\right)\sin ({137.5}^{\mathrm{o}}/2)}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\left(9.00\mathrm{x}{10}^{-2}\text{\hspace{0.17em}m}\right)\left(2.40\text{\hspace{0.17em}rad}\right)}\text{(}\because \mathrm{\theta }=2.40\text{\hspace{0.17em}rad}={137.5}^{\mathrm{o}}\text{)}\\ =5.39\text{N/C}\end{array}$

Hence, the magnitude of the electric field is $5.39\text{N/C}$.