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Suppose You Wanted to Hold Up an Electron Against the Force

question 29

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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k=1/4π( k = 1 / 4 \pi ε0=9.0×109 Nm2/C2,e=1.6×1019C,mproton =1.67×1027 kg,melectron =9.11×1031 kg) \left. \varepsilon _ { 0 } = 9.0 \times 10 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } , e = 1.6 \times 10 ^ { - 19 } \mathrm { C } , m _ { \text {proton } } = 1.67 \times 10 ^ { - 27 } \mathrm {~kg} , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} \right)


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