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Question Number 8823 by tawakalitu last updated on 30/Oct/16

When an electron is placed in an electric field, it experience an electric force whose magnitude is 1.6 times its weight . find the magnitude of the electric field.

$$\mathrm{When}\:\mathrm{an}\:\mathrm{electron}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{field}, \\ $$$$\mathrm{it}\:\mathrm{experience}\:\mathrm{an}\:\mathrm{electric}\:\mathrm{force}\:\mathrm{whose}\:\mathrm{magnitude} \\ $$$$\mathrm{is}\:\mathrm{1}.\mathrm{6}\:\mathrm{times}\:\mathrm{its}\:\mathrm{weight}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{magnitude} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{field}. \\ $$

Answered by sandy_suhendra last updated on 30/Oct/16

let F=electric force w=the weight of electron m_e =the electron mass=9.1×10^(−31) kg g=gravity=9.8 m/s^2 E=electric field e=elementary charge=1.6×10^(−19) C F=1.6w e.E=1.6 mg E= ((1.6 mg)/e) =((1.6×9.1×10^(−31) ×9.8)/(1.6×10^(−19) )) =8.9×10^(−11) N/C

$$\mathrm{let}\:\:\mathrm{F}=\mathrm{electric}\:\mathrm{force} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{w}=\mathrm{the}\:\mathrm{weight}\:\mathrm{of}\:\mathrm{electron} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{m}_{\mathrm{e}} =\mathrm{the}\:\mathrm{electron}\:\mathrm{mass}=\mathrm{9}.\mathrm{1}×\mathrm{10}^{−\mathrm{31}} \:\mathrm{kg} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{g}=\mathrm{gravity}=\mathrm{9}.\mathrm{8}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{E}=\mathrm{electric}\:\mathrm{field} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{e}=\mathrm{elementary}\:\mathrm{charge}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} \:\mathrm{C} \\ $$$$ \\ $$$$\mathrm{F}=\mathrm{1}.\mathrm{6w} \\ $$$$\mathrm{e}.\mathrm{E}=\mathrm{1}.\mathrm{6}\:\mathrm{mg} \\ $$$$\mathrm{E}=\:\frac{\mathrm{1}.\mathrm{6}\:\mathrm{mg}}{\mathrm{e}}\:=\frac{\mathrm{1}.\mathrm{6}×\mathrm{9}.\mathrm{1}×\mathrm{10}^{−\mathrm{31}} ×\mathrm{9}.\mathrm{8}}{\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} }\:=\mathrm{8}.\mathrm{9}×\mathrm{10}^{−\mathrm{11}} \:\mathrm{N}/\mathrm{C} \\ $$

Commented by tawakalitu last updated on 30/Oct/16

Thank you sir. God bless you.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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