A-charge-q-1-2-0-10-9-C-is-placed-at-the-point-x-0-y-4cm-and-another-charge-q-2-3-0-10-9-C-is-located-at-the-point-x-3cm-y-4cm-If-a-third-charge-q-3-4-0-10-9-C-is-placed-at-the-origi

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Question Number 40700 by Necxx last updated on 26/Jul/18

$$Acharge{q}_1=2.0\times {10}^{-9}Cisplaced$$$$atthepoint,(x=0,y=4cm)and$$$$anothercharge,q_2=-3.0\times {10}^{-9}C$$$$islocatedatthepoint,(x=3cm,y=4cm).$$$$Ifathirdcharge,q_3=4.0\times {10}^{-9}C$$$$isplacedattheorigin,$$$$a)obtainthexandycomponentof$$$$thetotalforceon{q}_3$$$$b)Calculatethemagnitudeand$$$$directionofthetotalforceon{q}_{3.}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

$$q_1=+ve{q}_2=-ve{q}_3=+ve$$$$soforceon{q}_{3}by{q}_{1}isrepulsivedirection$$$$-veyaxis$$$${\overrightarrow{F}}_1=9\times {10}^9\times \frac{2.0\times {10}^{-9}\times 4.0\times {10}^{-9}}{{(0.04)}^2}\times (-\overrightarrow{j})$$$$forceon{q}_{3}by{q}_{2}isattractive$$$${\overrightarrow{F}}_2=9\times {10}^9.\frac{3.0\times {10}^{-9}\times 4.0\times {10}^{-9}}{{(0.05)}^2}\left(\frac{0.03\overrightarrow{i}+0.04\overrightarrow{j}}{0.05}\right)$$$$netforce\overrightarrow{F}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$$$${\overrightarrow{F}}_1=\frac{72}{16}\times {10}^{9+4-18}(-\overrightarrow{j})=4.5\times {10}^{-5}(-\overrightarrow{j})$$$${\overrightarrow{F}}_2=\frac{108}{25}\times {10}^{9+4-18}\left(\frac{0.03\overrightarrow{i}+0.04\overrightarrow{j}}{0.05}\right)$$$${\overrightarrow{F}}_2=4.32\times {10}^{-5}(\frac{3}{5}i+\frac{4}{5}j)$$$$=2.592\times {10}^{-5}i+3.456\times {10}^{-5}j$$$$netforce$$$$\overrightarrow{F}=2.592\times {10}^{-5}i+7.956\times {10}^{-5}j$$$$miagnetude\approx \left(\sqrt{9+64}\right)\times {10}^{-5}=8.54\times {10}^{-5}N$$$$direction{tan}^{-1}\left(\frac{7.956}{2.592}\right)={tan}^{-1}(3.07)\dots$$

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