91影视

1. Radius of curvature of a circular rod,$\mathrm{R}=9.00\text{鈥塩尘}$
2. Uniformly distributed positive charge,$\mathrm{Q}=6.25\text{鈥塸颁}$
3. The subtended angle,$\mathrm{胃}=2.40\text{鈥塺补诲.}$

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,$\mathrm{E}=\frac{\mathrm{位蝉颈苍}(\mathrm{胃}/2)}{2{\mathrm{蟺蔚}}_{\mathrm{o}}\mathrm{R}}$ (i)

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

q = charge of the particle

胃 = the angle subtended that is expressed in radians

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

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

鈥淓lectric field of a charged circular rod鈥� 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:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{arc}}=\frac{(\left|\mathrm{Q}\right|/\mathrm{搁胃})\text{sin}(\mathrm{胃}/2)}{2{\mathrm{蟺蔚}}_{\mathrm{o}}\mathrm{R}}\\ =\frac{\left(\left|\mathrm{Q}\right|\right)\text{sin}(\mathrm{胃}/2)}{2{\mathrm{蟺蔚}}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{胃}}\\ =\frac{2\left(6.25\mathrm{x}{10}^{鈭�12}\text{鈥塁}\right)\sin ({137.5}^{\mathrm{o}}/2)}{4{\mathrm{蟺蔚}}_{\mathrm{o}}\left(9.00\mathrm{x}{10}^{鈭�2}\text{鈥尘}\right)\left(2.40\text{鈥塺补诲}\right)}\text{(}鈭�\mathrm{胃}=2.40\text{鈥塺补诲}={137.5}^{\mathrm{o}}\text{)}\\ =5.39\text{N/C}\end{array}$

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