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Question Number 60135 by Tawa1 last updated on 18/May/19

Answered by tanmay last updated on 18/May/19

$$q_1=8.5\times {10}^{-6}Cat{x}_1=3.0\times {10}^{-2}meter$$$$q_2=-21\times {10}^{-6}Cat{x}_2=9.0\times {10}^{-2}meter$$$$q_0(testcharge=1Cat{x}_o=6.0\times {10}^{-2}meter$$$$foceon{q}_{o}by{q}_{1}isrepulsivedirectedtowards+vexaxis$$$$F_1=9\times {10}^9\times \frac{8.5\times {10}^{-6}}{{(6\times {10}^{-2}-3\times {10}^{-2})}^2}$$$$F_1=9\times {10}^9\times \frac{8.5\times {10}^{-6}}{9\times {10}^{-4}}=8.5\times {10}^{7}N$$$$forceon{q}_{o}by{q}_{2}attractiveFdirected+vexaxis$$$$F_2=9\times {10}^9\times \frac{21\times {10}^{-6}}{{(9.0\times {10}^{-2}-6.0\times {10}^{-2})}^2}$$$$F_2=9\times {10}^9\times \frac{21\times {10}^{-6}}{9\times {10}^{-4}}=21\times {10}^{7}N$$$$sonetelectricfield=8.5\times {10}^7+21\times {10}^7$$$$=29.5\times {10}^{7}anddirection+vex[axis$$$$plscheck$$

Commented by Tawa1 last updated on 18/May/19

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

Commented by peter frank last updated on 19/May/19

$$thanks$$

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