„Elektromágneses terek alapjai - Szóbeli feladatok” változatai közötti eltérés
| 61. sor: | 61. sor: | ||
<math> | <math> | ||
U =- \int_{r_2}^{r_1} \vec{E}(r) d \vec{r} = - \int_{r_2}^{r_1} {q \over 2 \pi \varepsilon } | U =- \int_{r_2}^{r_1} \vec{E}(r) d \vec{r} = - \int_{r_2}^{r_1} {q \over 2 \pi \varepsilon } \cdot {1 \over r} dr = -{q \over 2 \pi \varepsilon } \cdot \left[ ln(r) \right]_{r_2}^{r_1} = {q \over {2\pi \varepsilon }}\ln {{{r_2}} \over {{r_1}}} | ||
</math> | </math> | ||
| 75. sor: | 75. sor: | ||
<math> | <math> | ||
C \buildrel \Delta \over = {Q \over U} = {{ql} \over U} \to C' = {C \over l} = {{{{ql} \over U}} \over l} = {q \over U} = { U {2 \pi \varepsilon \over ln{r_2 \over r_1}}} | C \buildrel \Delta \over = {Q \over U} = {{ql} \over U} \to C' = {C \over l} = {{{{ql} \over U}} \over l} = {q \over U} = { U {2 \pi \varepsilon \over ln{r_2 \over r_1}}} \cdot {1 \over U } = {{2\pi \varepsilon } \over {\ln {{{r_2}} \over {{r_1}}}}} | ||
</math> | </math> | ||
| 103. sor: | 103. sor: | ||
<math> | <math> | ||
G' = {{2\pi \sigma } \over {\ln {{{r_2}} \over {{r_1}}}}} = {1 \over R}{1 \over l} \to \sigma = {{\ln {{{r_2}} \over {{r_1}}}} \over {2\pi }}{1 \over R}{1 \over l} = {ln {6 \over 2} \over 2 \pi} | G' = {{2\pi \sigma } \over {\ln {{{r_2}} \over {{r_1}}}}} = {1 \over R}{1 \over l} \to \sigma = {{\ln {{{r_2}} \over {{r_1}}}} \over {2\pi }}{1 \over R}{1 \over l} = {ln {6 \over 2} \over 2 \pi} \cdot {1 \over 4 \cdot 10^6} \cdot {1 \over 200} \approx 218.6 \; {pS \over m} | ||
</math> | </math> | ||
}} | }} | ||