Question Number 60021 by Tawa1 last updated on 17/May/19
Answered by tanmay last updated on 17/May/19
![dW=F^→ .dr^→ =Fdrcoθ =(1/(4πε_0 ))((q_1 q_2 )/r^2 )drcosθ W=((q_1 q_2 cosθ)/(4πε_0 ))∫_r_1 ^r_2 (dr/r^2 ) W=((q_1 q_2 cosθ)/(4πε_0 ))×(−)[(1/r_2 )−(1/r_1 )] =((q_1 q_2 cosθ)/(4πε_0 ))×[(1/r_1 )−(1/r_2 )] now q_1 =2×10^(−4) C q_2 =3×10^(−5) C, θ=π r_1 =50×10^(−2) meter r_2 =20×10^(−2) meter (1/(4πε_0 ))=9×10^9 W=9×10^9 ×2×10^(−4) ×3×10^(−5) ×cosπ×[(1/(50×10^(−2) ))−(1/(20×10^(−2) ))] W=(−54)×[2−5] W=162](https://www.tinkutara.com/question/Q60023.png)
$${dW}=\overset{\rightarrow} {{F}}.{d}\overset{\rightarrow} {{r}} \\ $$$$\:\:\:\:\:\:\:\:\:={Fdrco}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{r}^{\mathrm{2}} }{drcos}\theta \\ $$$${W}=\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} {cos}\theta}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\int_{{r}_{\mathrm{1}} } ^{{r}_{\mathrm{2}} } \frac{{dr}}{{r}^{\mathrm{2}} } \\ $$$${W}=\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} {cos}\theta}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\left(−\right)\left[\frac{\mathrm{1}}{{r}_{\mathrm{2}} }−\frac{\mathrm{1}}{{r}_{\mathrm{1}} }\right] \\ $$$$=\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} {cos}\theta}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\left[\frac{\mathrm{1}}{{r}_{\mathrm{1}} }−\frac{\mathrm{1}}{{r}_{\mathrm{2}} }\right] \\ $$$${now}\:\:{q}_{\mathrm{1}} =\mathrm{2}×\mathrm{10}^{−\mathrm{4}} {C} \\ $$$${q}_{\mathrm{2}} =\mathrm{3}×\mathrm{10}^{−\mathrm{5}} {C},\:\:\:\:\theta=\pi \\ $$$${r}_{\mathrm{1}} =\mathrm{50}×\mathrm{10}^{−\mathrm{2}} {meter} \\ $$$${r}_{\mathrm{2}} =\mathrm{20}×\mathrm{10}^{−\mathrm{2}} {meter} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }=\mathrm{9}×\mathrm{10}^{\mathrm{9}} \\ $$$${W}=\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\mathrm{2}×\mathrm{10}^{−\mathrm{4}} ×\mathrm{3}×\mathrm{10}^{−\mathrm{5}} ×{cos}\pi×\left[\frac{\mathrm{1}}{\mathrm{50}×\mathrm{10}^{−\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{20}×\mathrm{10}^{−\mathrm{2}} }\right] \\ $$$${W}=\left(−\mathrm{54}\right)×\left[\mathrm{2}−\mathrm{5}\right] \\ $$$${W}=\mathrm{162} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$
Commented by Tawa1 last updated on 17/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$