En liknande del = RdO med laddning dg ger E-fält som en punktladdning (PH F=3.1)

|\vec{dE}| = \frac{dg}{4\pi\varepsilon_0 R^2} = dE \quad dE \text{ riktat enligt figur} dg = \lambda \cdot dE = \frac{Q}{R \cdot \frac{5\pi}{4}} \cdot RdO = \frac{4Q \, dO}{5\pi} x\text{-komp: } dE_x = dE(-\cos O) = -\frac{4Q \cos \theta \, dO}{4\pi\varepsilon_0 \cdot 5\pi R^2} y\text{-komp: } dE_y = dE(-\sin \theta) = -\frac{4Q \sin \theta \, dO}{4\pi\varepsilon_0 \cdot 5\pi R^2}

Integrera bidragen från $\theta = 0$ till $\theta = 5\pi/4$

E_x = \int_{\theta=0}^{\frac{5\pi}{4}} dE_x = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \int_{5\pi/4}^{\frac{5\pi}{4}} \cos \theta \, d\theta = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \left[ \sin \theta \right]_0^{5\pi/4} = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \left( -\frac{1}{\sqrt{2}} - 0 \right) = \frac{Q}{5\pi^2 \varepsilon_0 R^2 \sqrt{2}} \quad (E_x > 0) E_y = \int_{\theta=0}^{\frac{5\pi}{4}} dE_y = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \int_{0}^{\frac{5\pi}{4}} \sin \theta \, d\theta = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \left[ -\cos \theta \right]_0^{5\pi/4} = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \left( -(-\frac{1}{\sqrt{2}}) - (-1) \right) = -\frac{Q}{5\pi^2 \varepsilon_0 R^2} \left( \frac{1 + \sqrt{2}}{\sqrt{2}} \right) \quad (E_y < 0)

Svar:

\vec{E} = \frac{Q}{5\sqrt{2}\pi^2 \varepsilon_0 R^2} \left( \hat{i} - (1 + \sqrt{2}) \hat{j} \right)

(riktningen snett medelnd höger rimlig för

Q > 0

?)