91Ó°ÊÓ

The existence of a charged object affects the realm around it, causing an electrical field to create therein space. A charged object produces an electrical field, which could a change within the space or field surrounding it. The extraordinary modification of the realm would be felt by other charges in this field.

Small angle, $\mathrm{Δ}S$

Angle,$\mathrm{Δ}\mathrm{θ}=\frac{\mathrm{Δ}S}{R}$

Length, $L=10cm$

Charge,$Q=30\mathrm{nC}$

radius of the Semicircle, $R=\frac{L}{\mathrm{Ï€}}$

$\mathrm{Δ}\mathrm{θ}=\frac{\mathrm{Δ}S}{R}$

CALC Charge Q is evenly distributed throughout the length L of a thin, flexible rod. Figure $\text{P23.470}$ shows how the rod is bent into a semicircle. Graph $P23.47$ Cenier

a. Find the following expression for the electric field E at the semicircle's centre: As spans a little angle, a small amount of arc length $40=A5\sqrt{R}$, where R denotes the radius of the circle. .

$L=\mathrm{π}R$and $\mathrm{λ}=\frac{Q}{\mathrm{π}R}$, then

${\stackrel{→}{E}}_{ix}=E_i=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{\mathrm{Δ}Q}{r_i^2}$

${\stackrel{→}{E}}_{ix}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{\mathrm{λ}R\mathrm{Δ}\mathrm{θ}}{R^2}$

${\stackrel{→}{E}}_{ix}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{R\mathrm{Δ}Q}{R^2}×\frac{Q}{\mathrm{Ï€}R}$

${\stackrel{→}{E}}_{ix}=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{Q\mathrm{Δ}Q}{\mathrm{Ï€}{R}^2}$

Integrate equations, $\mathrm{θ}=\frac{\mathrm{π}}{2}$to $\mathrm{θ}=\frac{3\mathrm{π}}{2}$

$\stackrel{→}{E}={\stackrel{→}{E}}_{i-x}×\cos \mathrm{θ}i$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{Q\mathrm{Δ}Q}{\mathrm{Ï€}{R}^2}×\cos \mathrm{θ}$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{Q}{\mathrm{Ï€}{R}^2}{∫}_{\frac{\mathrm{Ï€}}{2}}^{\frac{3\mathrm{Ï€}}{2}}\cos \mathrm{θ}d\mathrm{θ}$

Imposing restrictions to the magnitude of the electrical field,

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×{\left.\frac{Q}{\mathrm{Ï€}{R}^2}[-\sin \mathrm{θ}]\right|}_{\frac{\mathrm{Ï€}}{2}}^{\frac{3\mathrm{Ï€}}{2}}$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{Q}{\mathrm{Ï€}{R}^2}\left[-\sin \left(\frac{3\mathrm{Ï€}}{2}\right)+\left(-\sin \left(\frac{\mathrm{Ï€}}{2}\right)\right)\right]$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{Q}{\mathrm{Ï€}{R}^2}×2$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{2Q}{\mathrm{Ï€}{R}^2}$

$\stackrel{→}{E}={\stackrel{→}{E}}_{i-x}×\cos \mathrm{θ}i$

If $L=10cm$ and $Q=30nC$, calculate the field strength.

$R=\frac{L}{\mathrm{Ï€}}$, then $L=R\mathrm{Ï€}$

$\frac{Q}{\mathrm{π}\left(\frac{L^2}{{\mathrm{λ}}^2}\right)}$$=\frac{Q\mathrm{π}}{L^2}$

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{2\mathrm{Ï€}Q}{L^2}$

Given $L=0.1m$and charge $30×{10}^{-9}\mathrm{C}$, solve equation

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{2Q\mathrm{λ}}{L^2}$

$=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}×\frac{\left(30×{10}^{-4}\mathrm{C}\right)\left(2\mathrm{λ}\right)}{(0.1\mathrm{m}{)}^2}$

$=170×{10}^5\mathrm{N}/\mathrm{C}$