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Question-13831

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Question Number 13831 by tawa tawa last updated on 24/May/17
Answered by ajfour last updated on 24/May/17
(5). F_e =(e^2 /(4πε_0 r^2 )) ; F_g =G((m_e m_p )/r^2 ) (F_e /F_g )=((e^2 /(m_e m_p )))((1/(4πε_0 G))) =(((1.6×10^(−19) )^2 (9×10^9 ))/((9.1×10^(−31) )(1.7×10^(−27) )(6.67×10^(−11) ))) =((2.56×9)/(9.1×1.7×6.67))×10^(−38+9+31+27+11) ≈(((5/2))/((5/3)(20/3)))×10^(40) ≈(9/(40))×10^(40) or =2.2×10^(39) . (6). α particle has charge q=+2e F=(((2e)(e))/(4πε_0 r^2 )) F=((2(1.6×10^(−19) )^2 (9×10^9 ))/((10^(−13) )^2 )) =2×2.56×9×10^(−38+9+26) =5.12×9×10^(−3) N F=4.608×10^(−2) N ≈ 0.46mN .
$$\left(\mathrm{5}\right). \\ $$$${F}_{{e}} =\frac{{e}^{\mathrm{2}} }{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} }\:\:\:;\:\:{F}_{{g}} ={G}\frac{{m}_{{e}} {m}_{{p}} }{{r}^{\mathrm{2}} } \\ $$$$\frac{{F}_{{e}} }{{F}_{{g}} }=\left(\frac{{e}^{\mathrm{2}} }{{m}_{{e}} {m}_{{p}} }\right)\left(\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {G}}\right) \\ $$$$\:\:\:\:\:=\frac{\left(\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} \right)^{\mathrm{2}} \left(\mathrm{9}×\mathrm{10}^{\mathrm{9}} \right)}{\left(\mathrm{9}.\mathrm{1}×\mathrm{10}^{−\mathrm{31}} \right)\left(\mathrm{1}.\mathrm{7}×\mathrm{10}^{−\mathrm{27}} \right)\left(\mathrm{6}.\mathrm{67}×\mathrm{10}^{−\mathrm{11}} \right)} \\ $$$$\:\:\:\:=\frac{\mathrm{2}.\mathrm{56}×\mathrm{9}}{\mathrm{9}.\mathrm{1}×\mathrm{1}.\mathrm{7}×\mathrm{6}.\mathrm{67}}×\mathrm{10}^{−\mathrm{38}+\mathrm{9}+\mathrm{31}+\mathrm{27}+\mathrm{11}} \\ $$$$\:\:\:\:\approx\frac{\left(\mathrm{5}/\mathrm{2}\right)}{\left(\mathrm{5}/\mathrm{3}\right)\left(\mathrm{20}/\mathrm{3}\right)}×\mathrm{10}^{\mathrm{40}} \\ $$$$\:\:\:\approx\frac{\mathrm{9}}{\mathrm{40}}×\mathrm{10}^{\mathrm{40}} \:\:{or}\:\:=\mathrm{2}.\mathrm{2}×\mathrm{10}^{\mathrm{39}} \:. \\ $$$$\left(\mathrm{6}\right). \\ $$$$\alpha\:{particle}\:{has}\:{charge}\:{q}=+\mathrm{2}{e} \\ $$$${F}=\frac{\left(\mathrm{2}{e}\right)\left({e}\right)}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {r}^{\mathrm{2}} } \\ $$$${F}=\frac{\mathrm{2}\left(\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} \right)^{\mathrm{2}} \left(\mathrm{9}×\mathrm{10}^{\mathrm{9}} \right)}{\left(\mathrm{10}^{−\mathrm{13}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:=\mathrm{2}×\mathrm{2}.\mathrm{56}×\mathrm{9}×\mathrm{10}^{−\mathrm{38}+\mathrm{9}+\mathrm{26}} \\ $$$$\:\:\:=\mathrm{5}.\mathrm{12}×\mathrm{9}×\mathrm{10}^{−\mathrm{3}} {N} \\ $$$${F}=\mathrm{4}.\mathrm{608}×\mathrm{10}^{−\mathrm{2}} {N}\:\approx\:\mathrm{0}.\mathrm{46}{mN}\:. \\ $$
Commented by tawa tawa last updated on 24/May/17
God will always help you too. Thank you sir.
$$\mathrm{God}\:\mathrm{will}\:\mathrm{always}\:\mathrm{help}\:\mathrm{you}\:\mathrm{too}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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