13. Lösungen der Übungsaufgaben

$$N=\frac{M}{m}=\frac{10^{-3}}{2{,}9\cdot 1{,}67\cdot 10^{-27}}=2{,}06\cdot 10^{22}$$

$$\begin{aligned}\displaystyle Q&\displaystyle=+e\cdot 0{,}1\cdot 2{,}6\cdot 10^{22}\\ \displaystyle&\displaystyle=2{,}6\cdot 10^{21}\cdot 1{,}6\cdot 10^{-19}\,\mathrm{C}\\ \displaystyle&\displaystyle=4{,}16\cdot 10^{2}\,\mathrm{C}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle V&\displaystyle=\frac{m}{\varrho}=1{,}03\,\mathrm{cm^{3}}=\frac{4}{3}\pi r^{3}\\ \displaystyle\Rightarrow r&\displaystyle=\left(\frac{3\cdot 1{,}03}{4\pi}\right)^{1/3}\mathrm{cm}=0{,}63\,\mathrm{cm}\;,\end{aligned}$$

$$S=4\pi r^{2}=4{,}93\,\mathrm{cm^{2}},$$

$$\sigma=\frac{Q}{4\pi r^{2}}=8{,}4\cdot 10^{5}\,\mathrm{C/m^{2}}\;,$$

$$F_{C}=\frac{1}{4\pi\varepsilon_{0}}\frac{Q^{2}}{r^{2}}=1{,}56\cdot 10^{15}\,\mathrm{N}.$$

$$E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}=9{,}6\cdot 10^{16}\,\mathrm{V/m}\;.$$

$$\begin{aligned}\displaystyle&\displaystyle\Rightarrow\;{\operatorname{tan}}(\varphi/2)=\frac{F_{\text{el}}}{m\cdot g}\;.\\ \displaystyle F_{\text{el}}&\displaystyle=\frac{Q^{2}}{4\pi\varepsilon_{0}(2L\cdot\sin\varphi/2)^{2}}\\ \displaystyle&\displaystyle\Rightarrow\;\frac{\sin^{3}(\varphi/2)}{\cos(\varphi/2)}=\frac{Q^{2}}{16\pi\varepsilon_{0}L^{2}\cdot mg}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle&\displaystyle\Rightarrow\;\frac{\sin^{3}(\varphi/2)}{\cos(\varphi/2)}=\frac{10^{-16}}{16\pi\varepsilon_{0}\cdot 10^{-2}\cdot 9{,}81}=2{,}3\cdot 10^{-6}\\ \displaystyle&\displaystyle\Rightarrow\;\sin\varphi\approx\varphi,\;\cos\varphi\approx 1\\ \displaystyle&\displaystyle\Rightarrow\;\varphi\approx 2\cdot\sqrt[3]{2{,}3\cdot 10^{-6}}=2{,}6\cdot 10^{-2}\,\mathrm{rad}\approx 1{,}5^{\circ}\\ \displaystyle&\displaystyle\Rightarrow\;\text{Abstand}\;r=2L\cdot\sin\varphi/2\\ \displaystyle&\displaystyle=0{,}026\,\mathrm{m}=2{,}6\,\mathrm{cm}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle\boldsymbol{E}&\displaystyle=\frac{\sigma}{2\varepsilon_{0}}\hat{\boldsymbol{x}}\;\Rightarrow\;\boldsymbol{F}_{\text{el}}=\frac{Q\cdot\sigma}{2\varepsilon_{0}}\hat{\boldsymbol{x}}\\ \displaystyle\Rightarrow\;{\operatorname{tan}}\varphi&\displaystyle=\frac{F_{\text{el}}}{m\cdot g}=\frac{Q\cdot\sigma}{2\varepsilon_{0}m\cdot g}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle\Rightarrow\;{\operatorname{tan}}\varphi&\displaystyle=\frac{10^{-8}\cdot 1{,}5\cdot 10^{-5}}{2\cdot 8{,}85\cdot 10^{-12}\cdot 0{,}05\cdot 9{,}81}\\ \displaystyle&\displaystyle=1{,}7\cdot 10^{-2}\\ \displaystyle\Rightarrow\;\varphi&\displaystyle=1^{\circ}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle F&\displaystyle=\int\limits_{\alpha=\alpha_{\text{i}}}^{\alpha_{\text{a}}}\frac{q\cdot\sigma}{2\varepsilon_{0}}\sin\alpha\cdot\,\mathrm{d}\alpha\\ \displaystyle&\displaystyle=\frac{q\cdot\sigma}{2\varepsilon_{0}}(\cos\alpha_{\text{i}}-\cos\alpha_{\text{a}})\;,\\ \displaystyle\cos\alpha&\displaystyle=\frac{x}{\sqrt{r^{2}+x^{2}}}\\ \displaystyle\Rightarrow\;F&\displaystyle=\frac{q\,\sigma\,x}{2\varepsilon_{0}}\left[\frac{1}{\sqrt{R_{\text{i}}^{2}+x^{2}}}-\frac{1}{\sqrt{R_{\text{a}}^{2}+x^{2}}}\right]\;.\end{aligned}$$

$$F=\frac{q\cdot\sigma}{2\varepsilon_{0}}\left[1-\frac{x}{\sqrt{R_{\text{a}}^{2}+x^{2}}}\right]\;,$$

$$F=\frac{q\cdot\sigma}{2\varepsilon_{0}}\frac{1}{\sqrt{1+R_{\text{i}}^{2}/x^{2}}}\;,$$

$$F=\frac{q\cdot\sigma}{2\varepsilon_{0}}\;.$$

$$\phi(R)=\frac{Q}{4\pi\varepsilon_{0}R}\;.$$

$$\begin{aligned}\displaystyle&\displaystyle\Rightarrow\;\phi_{1}(R_{1})=\frac{Q_{1}}{4\pi\varepsilon_{0}R_{1}}=\phi_{2}(R_{2})=\frac{Q_{2}}{4\pi\varepsilon_{0}R_{2}}\\ \displaystyle&\displaystyle\Rightarrow\;\frac{Q_{1}}{Q_{2}}=\frac{R_{1}}{R_{2}}\;,\quad Q=Q_{1}+Q_{2}=Q_{1}\left(1+\frac{R_{2}}{R_{1}}\right)\\ \displaystyle&\displaystyle\Rightarrow\;Q_{1}=\frac{Q\cdot R_{1}}{R_{1}+R_{2}}\;,\quad Q_{2}=\frac{R_{2}\cdot Q}{R_{1}+R_{2}}\\ \displaystyle&\displaystyle E_{1}=\frac{Q_{1}}{4\pi\varepsilon_{0}R_{1}^{2}}=\frac{Q}{4\pi\varepsilon_{0}R_{1}(R_{1}+R_{2})}\\ \displaystyle&\displaystyle E_{2}=\frac{Q_{2}}{4\pi\varepsilon_{0}R_{2}^{2}}=\frac{Q}{4\pi\varepsilon_{0}R_{2}(R_{1}+R_{2})}\;.\end{aligned}$$

$$\phi(R)=\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right)\;.$$

$$\begin{aligned}\displaystyle\phi(R)&\displaystyle\approx\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{1}{R-a\cos\vartheta}\right.\\ \displaystyle&\displaystyle\left.\qquad\qquad\qquad\quad+\frac{1}{R+a\cos\vartheta}\right)\quad\text{f{\"u}r}\quad R\gg a\\ \displaystyle&\displaystyle=\frac{2Q}{4\pi\varepsilon_{0}R}\cdot\frac{1}{1-\frac{a^{2}}{R^{2}}\cos^{2}\vartheta}\;.\end{aligned}$$

$$\frac{1}{1-x}\approx 1+x+x^{2}+\cdots+x^{n}$$

$$\begin{aligned}\displaystyle\Rightarrow\;\phi(R)&\displaystyle=\frac{2Q}{4\pi\varepsilon_{0}R}\left(1+\frac{a^{2}}{R^{2}}\cos^{2}\vartheta\right.\\ \displaystyle&\displaystyle\qquad\qquad\!\qquad\left.+\frac{a^{4}}{R^{4}}\cos^{4}\vartheta+\cdots\right)\;.\end{aligned}$$

$$\boldsymbol{F}=-q\cdot{\mathop{\mathbf{grad}}}\,\phi(R)\;.$$

$$\begin{aligned}\displaystyle\phi&\displaystyle=\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\\ \displaystyle&\displaystyle=\frac{2aQ}{4\pi\varepsilon_{0}}\frac{\cos\vartheta}{R^{2}-a^{2}\cos^{2}\vartheta}\\ \displaystyle&\displaystyle=\frac{2|\,\boldsymbol{p}|\cdot\cos\vartheta}{4\pi\varepsilon_{0}R^{2}}\frac{1}{1-\frac{a^{2}}{R^{2}}\cos^{2}\vartheta}\\ \displaystyle&\displaystyle=\frac{2p\cdot\cos\vartheta}{4\pi\varepsilon_{0}R^{2}}\!\left[1+\frac{a^{2}}{R^{2}}\cos^{2}\vartheta+\frac{a^{4}}{R^{4}}\cos^{4}\vartheta+\cdots\right]\!.\end{aligned}$$

$$\boldsymbol{F}=\frac{2Qq}{4\pi\varepsilon_{0}R^{2}}\hat{\boldsymbol{R}}\;,$$

$$=-\frac{Q^{2}}{4\pi\varepsilon_{0}a}$$

$$=\frac{Q^{2}}{4\pi\varepsilon_{0}a}(-4+\sqrt{2})\approx-\frac{2{,}6Q^{2}}{4\pi\varepsilon_{0}a}$$

$$\begin{aligned}\displaystyle Q_{1}&\displaystyle=+Q:\;\left(\frac{-a}{2},0\right);\\ \displaystyle Q_{2}&\displaystyle=+Q:\;\left(\frac{+a}{2},0\right);\\ \displaystyle r_{1}^{2}&\displaystyle=r_{2}^{2}=\frac{a^{2}}{4}\,;\\ \displaystyle Q_{3}&\displaystyle=-Q:\;\left(0,\frac{a}{2}\sqrt{3}\right);\\ \displaystyle Q_{4}&\displaystyle=-Q:\;\left(0,\frac{-a}{2}\sqrt{3}\right);\\ \displaystyle r_{3}^{2}&\displaystyle=r_{4}^{2}=\frac{3a^{2}}{4}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle QM_{xx}&\displaystyle=Q_{1}\left(\frac{3}{4}a^{2}-\frac{1}{4}a^{2}\right)+Q_{2}\left(\frac{3}{4}a^{2}-\frac{1}{4}a^{2}\right)\\ \displaystyle&\displaystyle\quad+Q_{3}\left(-\frac{3}{4}a^{2}\right)+Q_{4}\left(-\frac{3}{4}a^{2}\right)\\ \displaystyle&\displaystyle=\frac{5}{2}Qa^{2},\\ \displaystyle QM_{yy}&\displaystyle=-\frac{7}{2}Qa^{2};\quad QM_{zz}=+a^{2}Q;\\ \displaystyle QM_{xy}&\displaystyle=QM_{xz}=QM_{yz}=0\;.\end{aligned}$$

$$\begin{aligned}\displaystyle Q_{1}&\displaystyle=-Q:\;(-a,0)\;;\quad Q_{2}=2Q:\;(0,0)\;;\\ \displaystyle Q_{3}&\displaystyle=-Q:\;(+a,0)\;.\end{aligned}$$

$$\begin{aligned}\displaystyle QM_{xx}&\displaystyle=-4Qa^{2};\quad QM_{yy}=2Qa^{2};\\ \displaystyle QM_{zz}&\displaystyle=2Qa^{2};\\ \displaystyle QM_{xy}&\displaystyle=QM_{xz}=QM_{yz}=0\;.\end{aligned}$$

$$\begin{aligned}\displaystyle\phi(r)&\displaystyle=\frac{Q}{8\pi\varepsilon_{0}R^{3}}(3R^{2}-r^{2})\quad\text{mit}\;Q=\frac{4}{3}\pi\varrho_{\text{el}}R^{3}\\ \displaystyle&\displaystyle=\frac{1}{6}\frac{\varrho_{\text{el}}}{\varepsilon_{0}}(3R^{2}-r^{2})\;.\end{aligned}$$

$$\phi(r)=\frac{Q}{4\pi\varepsilon_{0}r}\;.$$

$$\begin{aligned}\displaystyle W_{1}&\displaystyle=q\cdot\big[\phi(R)-\phi(0)\big]=\frac{q\cdot Q}{4\pi\varepsilon_{0}R}\cdot\left(1-\frac{3}{2}\right)\\ \displaystyle&\displaystyle=-\frac{q\cdot Q}{8\pi\varepsilon_{0}R}\;.\end{aligned}$$

$$W_{2}=-\frac{q\cdot Q}{4\pi\varepsilon_{0}R}\;,$$

$$\begin{aligned}\displaystyle&\displaystyle E(r)=-\frac{\mathrm{d}\phi(r)}{\mathrm{d}r}:\\ \displaystyle&\displaystyle{\text{a)}}\;\displaystyle\boldsymbol{E}(r)=\frac{Q\cdot r}{4\pi\varepsilon_{0}R^{3}}\hat{\boldsymbol{r}}\quad\text{f{\"u}r}\quad r\leq R,\\ \displaystyle&\displaystyle{\text{b)}}\;\displaystyle\boldsymbol{E}(r)=\frac{Q}{4\pi\varepsilon_{0}r^{2}}\hat{\boldsymbol{r}}\quad\text{f{\"u}r}\quad r\geq R.\end{aligned}$$

$$\begin{aligned}\displaystyle\frac{1}{|\boldsymbol{R}-\boldsymbol{r}|}&\displaystyle=\frac{1}{R}-\left(x\frac{\partial}{\partial X}\frac{1}{R}+y\frac{\partial}{\partial Y}\frac{1}{R}+z\frac{\partial}{\partial Z}\frac{1}{R}\right)\\ \displaystyle&\displaystyle\quad+\frac{1}{2}\left[xx\frac{\partial^{2}}{\partial X^{2}}\frac{1}{R}+xy\frac{\partial^{2}}{\partial X\partial Y}\frac{1}{R}\right.\\ \displaystyle&\displaystyle\quad+\left.\cdots+zz\frac{\partial^{2}}{\partial Z^{2}}\frac{1}{R}\right]+\cdots.\\ \displaystyle R&\displaystyle=\sqrt{X^{2}+Y^{2}+Z^{2}}\;\Rightarrow\;\frac{\partial}{\partial X}\frac{1}{R}=\frac{-X}{R^{3}}\\ \displaystyle\frac{\partial^{2}}{\partial X\partial Y}&\displaystyle\frac{1}{R}=\frac{3}{2}\frac{X\cdot Y}{R^{5}}\quad\text{etc}.\end{aligned}$$

$$\begin{aligned}\displaystyle\phi(R)&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\sum\frac{Q_{i}}{|\boldsymbol{R}-\boldsymbol{r_{i}}|}\\ \displaystyle&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\left[\frac{\sum Q_{i}}{R}+\frac{1}{R^{3}}\sum(Q_{i}\boldsymbol{r_{i}})\cdot\boldsymbol{R}\right.\\ \displaystyle&\displaystyle\quad+\frac{1}{R^{5}}\frac{1}{2}\sum Q_{i}\big\{\left(3x_{i}^{2}-r_{i}^{2}\right)X^{2}\\ \displaystyle&\displaystyle\quad+\left(3y_{i}^{2}-r_{i}^{2}\right)Y^{2}+\left(3z_{i}^{2}-r_{i}^{2}\right)Z^{2}\\ \displaystyle&\displaystyle\quad+2\cdot 3x_{i}y_{i}XY+2\cdot 3y_{i}z_{i}YZ\\ \displaystyle&\displaystyle\quad+2\cdot 3x_{i}z_{i}XZ\big\}\bigg]\;.\end{aligned}$$

$$\phi(R)=\frac{1}{4\pi\varepsilon_{0}}\int\limits_{V}\frac{\varrho_{\text{el}}}{|\boldsymbol{R}-\boldsymbol{r}|}\,\mathrm{d}V=\frac{Q}{4\pi\varepsilon_{0}R}\;.$$

$$r=\sqrt{y^{2}+(R-x)^{2}}$$

$$\mathrm{d}\phi=\frac{\mathrm{d}Q}{4\pi\varepsilon_{0}r}=\frac{\varrho_{\text{el}}}{2\varepsilon_{0}}\frac{y\,\mathrm{d}y}{\sqrt{y^{2}+(R-x)^{2}}}\,\mathrm{d}x\;.$$

$$\phi_{\text{Scheibe}}=\frac{\varrho_{\text{el}}}{2\varepsilon_{0}}\left[\int\limits_{y=0}^{\sqrt{a^{2}-x^{2}}}\frac{y\,\mathrm{d}y}{\sqrt{y^{2}-(R-x)^{2}}}\right]\,\mathrm{d}x\;.$$

$$\phi_{\text{Kugel}}=\frac{\varrho_{\text{el}}}{\varepsilon_{0}}\frac{a^{3}}{3R}=\frac{Q}{4\pi\varepsilon_{0}R}\;.$$

$$\boldsymbol{E}=\frac{\lambda}{2\pi\varepsilon_{0}r}\hat{\boldsymbol{r}}$$

$$\begin{aligned}\displaystyle\boldsymbol{E}&\displaystyle=\{E_{x},0,0\}\\ \displaystyle E_{x}&\displaystyle=\frac{\lambda}{2\pi\varepsilon_{0}}\left[\frac{1}{a+x}-\frac{1}{a-x}+\frac{2x}{a^{2}+x^{2}}\right]\\ \displaystyle&\displaystyle=\frac{\lambda}{2\pi\varepsilon_{0}}\left[\frac{-2x}{a^{2}-x^{2}}+\frac{2x}{a^{2}+x^{2}}\right]\\ \displaystyle&\displaystyle=-\frac{\lambda}{\pi\varepsilon_{0}}\frac{2x^{3}}{a^{4}-x^{4}}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle E_{x}&\displaystyle=-\frac{\lambda}{\pi\varepsilon_{0}}\frac{2(a-R)^{3}}{a^{4}-(a-R)^{4}}\\ \displaystyle&\displaystyle=-\frac{\lambda}{2\pi\varepsilon_{0}R}\cdot\frac{4\cdot R(a-R)^{3}}{a^{4}-(a-R)^{4}}\;.\end{aligned}$$

$$E_{x}=-\frac{\lambda}{2\pi\varepsilon_{0}R}\cdot 0{,}8\,\mathrm{V/m}\;.$$

$$E=\frac{\lambda}{2\pi\varepsilon_{0}R}\cdot\frac{4R(a+R)^{3}}{(a+R)^{4}-a^{4}}=\frac{\lambda}{2\pi\varepsilon_{0}R}\cdot 1{,}18\,\mathrm{V/m}$$

$$\begin{aligned}\displaystyle C&\displaystyle=\varepsilon_{0}\cdot\frac{A}{d}=\frac{8{,}85\cdot 10^{-12}\cdot 0{,}1}{0{,}01}\mathrm{F}\\ \displaystyle&\displaystyle=8{,}85\cdot 10^{-11}\,\mathrm{F}=88{,}5\,\mathrm{pF}\;;\\ \displaystyle Q&\displaystyle=C\cdot U=8{,}85\cdot 10^{-11}\cdot 5\cdot 10^{3}\,\mathrm{C}\\ \displaystyle&\displaystyle=4{,}4\cdot 10^{-7}\,\mathrm{C}\;;\end{aligned}$$

$$E=\frac{U}{d}=5\cdot 10^{5}\,\mathrm{V/m}\;.$$

$$W=\int\limits_{0}^{\infty}I^{2}\cdot R\cdot\,\mathrm{d}t\;.$$

$$\begin{aligned}\displaystyle W&\displaystyle=\frac{U_{0}^{2}}{R}\cdot\left(-\frac{R\cdot C}{2}\right)\cdot\left.{\text{e}}^{-2t/(R{}C)}\right|_{0}^{\infty}\\ \displaystyle&\displaystyle=\frac{U_{0}^{2}C}{2}\;.\end{aligned}$$

$$\begin{aligned}\displaystyle\Rightarrow\;|\boldsymbol{D}|&\displaystyle=1{,}6\cdot 10^{-19}\cdot 5\cdot 10^{-11}\cdot 5\cdot 10^{5}\,\mathrm{N\cdot m}\\ \displaystyle&\displaystyle=4\cdot 10^{-24}\,\mathrm{N\cdot m}\;,\\ \displaystyle W_{\text{pot}}&\displaystyle=\boldsymbol{p}\cdot\boldsymbol{E}=4\cdot 10^{-24}\,\mathrm{N\cdot m}\;.\end{aligned}$$

$$\frac{1}{C_{\text{g}}}=\frac{1}{C}+\frac{1}{3C}\;\Rightarrow\;C_{\text{g}}=\frac{3}{4}C\;.$$

$$C+\frac{1}{2}C=\frac{3}{2}C\;.$$

$$\frac{1}{C}+\frac{1}{3/2C}=\frac{5}{3C}\Rightarrow C_{r}=\frac{3}{5}C\;.$$

$$\frac{3}{5}C\!+2C=\frac{13}{5}C\Rightarrow\;\text{Gesamtkapazit{\"a}t\;ist\;}\frac{13}{18}C\;.$$

$$E=\frac{U}{d}=\frac{3}{4}\,\frac{Q}{C\cdot d}\;.$$

$$\Delta\phi =\frac{1}{R}\frac{\partial}{\partial R}\left(R\cdot\frac{\partial\phi}{\partial R}\right)=0$$

$$\Rightarrow\;\phi =c_{1}\ln R+c_{2}$$

$$c_{2} =\phi_{1}-c_{1}\ln R_{1}\;,$$

$$c_{1} =\frac{\phi_{2}-\phi_{1}}{\ln(R_{2}/R_{1})}$$

$$\Rightarrow\;\phi(R) =\phi_{1}+\frac{\phi_{2}-\phi_{1}}{\ln(R_{2}/R_{2})}\ln(R/R_{1})\;,$$

$$E(R) =-\frac{\partial\phi}{\partial R}=\frac{-(\phi_{2}-\phi_{1})}{\ln(R_{2}/R_{1})}\frac{1}{R}\;.$$

$$\begin{aligned}\displaystyle\frac{mv_{0}^{2}}{R_{0}}&\displaystyle=e\cdot E(R_{0})=\frac{2e}{R_{1}+R_{2}}\frac{\phi_{1}-\phi_{2}}{\ln(R_{2}/R_{1})}\\ \displaystyle\Rightarrow\;U&\displaystyle=\frac{R_{1}+R_{2}}{2e}\ln(R_{2}/R_{1})\cdot\frac{m}{R}v_{0}^{2}\\ \displaystyle&\displaystyle=\frac{R_{1}+R_{2}}{2R}\frac{m}{e}v_{0}^{2}\ln\frac{R_{2}}{R_{1}}\;.\end{aligned}$$

$$U=\frac{m}{e}v_{0}^{2}\ln\frac{R_{2}}{R_{1}}\;.$$

$$v\cdot R=v_{0}\cdot R_{0}={\mathrm{konst.}}\;.$$

$$m\cdot\delta\,\ddot{R}-m\cdot\frac{v^{2}}{R}-e\cdot E(R_{0}+\delta R)=0\;.$$

$$E(R_{0}+\delta R)=E(R_{0})+\left(\frac{\mathrm{d}E}{\mathrm{d}R}\right)_{R_{0}}\delta R+\cdots\;.$$

$$\frac{\mathrm{d}E}{\mathrm{d}R}=\frac{U}{\ln(R_{2}/R_{1})}\frac{1}{R^{2}}\;.$$

$$\delta\,\ddot{R}-\frac{v_{0}^{2}}{R^{3}}R_{0}^{2}+\frac{v_{0}^{2}}{R_{0}}\left(1-\frac{\delta R}{R_{0}}\right)=0\;.$$

$$\begin{aligned}\displaystyle&\displaystyle\frac{1}{R^{3}}=\frac{1}{R_{0}^{3}\left(1+\frac{\delta R}{R_{0}}\right)^{3}}\approx\frac{1}{R_{0}^{3}}-\frac{3}{R_{0}^{4}}\delta R+\dots\\ \displaystyle&\displaystyle\Rightarrow\;\delta\,\ddot{R}-\frac{v_{0}^{2}}{R_{0}}\left(1-3\frac{\delta R}{R_{0}}-1+\frac{\delta R}{R_{0}}\right)=0\\ \displaystyle&\displaystyle\Rightarrow\;\delta\,\ddot{R}+2\omega_{0}^{2}\delta R=0\quad\text{mit}\quad\omega_{0}=\frac{v_{0}}{R_{0}}\;.\end{aligned}$$

$$\delta R=R_{0}\cdot\sin\left[\sqrt{2}\omega_{0}\cdot t\right]\;,$$

$$\mathrm{d}\boldsymbol{E}=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda\cdot\,\mathrm{d}L}{R^{2}}\{\cos\varphi,\sin\varphi,0\}$$

$$\begin{aligned}\displaystyle E_{x}&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda}{R^{2}}\int\limits_{\varphi_{1}}^{\varphi_{2}}R\cdot\cos\varphi\,\mathrm{d}\varphi\;,\\ \displaystyle E_{y}&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda}{R^{2}}\int\limits_{\varphi_{1}}^{\varphi_{2}}R\cdot\sin\varphi\,\mathrm{d}\varphi\;,\\ \displaystyle\varphi_{1}&\displaystyle=\frac{\pi}{2}-\frac{\alpha}{2}=\frac{\pi}{2}-\frac{L}{2R}\;,\\ \displaystyle\varphi_{2}&\displaystyle=\frac{\pi}{2}+\frac{L}{2R}\\ \displaystyle\Rightarrow\;E_{x}&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda}{R}(\sin\varphi_{2}-\sin\varphi_{1})\\ \displaystyle&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda}{R}\left(\cos\frac{L}{2R}-\cos\frac{L}{2R}\right)=0\;,\\ \displaystyle E_{y}&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{\lambda}{R}(\cos\varphi_{1}-\cos\varphi_{2})\\ \displaystyle&\displaystyle=\frac{1}{4\pi\varepsilon_{0}}\frac{2\lambda}{R}\,\sin\frac{L}{2R}\;.\end{aligned}$$

$$|\boldsymbol{E}|=\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{R}\sin\frac{L}{2R}\;.$$