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Question Number 60021 by Tawa1 last updated on 17/May/19

Answered by tanmay last updated on 17/May/19

$$dW=\overrightarrow{F}.d\overrightarrow{r}$$$$=Fdrco\theta$$$$=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}drcos\theta$$$$W=\frac{q_1{q}_{2}cos\theta }{4\pi {ϵ}_0}{\int }_{r_1}^{r_2}\frac{dr}{r^2}$$$$W=\frac{q_1{q}_{2}cos\theta }{4\pi {ϵ}_0}\times (-)[\frac{1}{r_2}-\frac{1}{r_1}]$$$$=\frac{q_1{q}_{2}cos\theta }{4\pi {ϵ}_0}\times [\frac{1}{r_1}-\frac{1}{r_2}]$$$$now{q}_1=2\times {10}^{-4}C$$$$q_2=3\times {10}^{-5}C,\theta =\pi$$$$r_1=50\times {10}^{-2}meter$$$$r_2=20\times {10}^{-2}meter$$$$\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$$$$W=9\times {10}^9\times 2\times {10}^{-4}\times 3\times {10}^{-5}\times cos\pi \times [\frac{1}{50\times {10}^{-2}}-\frac{1}{20\times {10}^{-2}}]$$$$W=(-54)\times [2-5]$$$$W=162$$$$$$

Commented by Tawa1 last updated on 17/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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