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Calculate the ratio of the electrostatic of gravitational inteaction forces between two electrons, between two protons. At what values of the specific cahreg `q//m` of a praticle would these forces become equal (in their absoule values) in the case of interaction of indentical) particles ? |
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Answer» `F_(el)` (for electrons) = `(q^(2))/(4pi epsilon_(0) r^(2))` and `F_(gr) = (gamma m^(2))/(r^(2))` Thus, `(F_(el))/(F_(gr))` for electrons) `= (q^(2))/(4pi epsilon_(0) gamma m^(2))` `= ((1.602xx10^(-19) C)^(2))/(((1)/(9xx10^(9))) xx6.67xx10^(-11) m^(3)//(kg . s^(2)) xx (9.11xx10^(-31) kg)^(2)) = 4xx10^(42)` Similarly `(F_(el))/(F_(gr))` (for proton) `= (q^(2))/(4pi epsilon_(0) gamma m^(2))` `= ((1.602xx10^(-19) C)^(2))/(((1)/(9xx10^(9))) xx6.67xx10^(-11) m^(3)//(kg . s^(2)) xx (1.672xx10^(-27) kg)^(2)) = 1xx10^(36)` For `F_(el) = F_(gr)` `(q^(2))/(4pi epsilon_(0) r^(2))` or `(gamma m^(2))/(r^(2))` or `(q)/(m) = sqrt(4pi epsilon_(0) gamma)` `= sqrt((6.67xx10^(-11) m^(3) (kg-s^(2)))/(9xx10^(9))) = 0.86xx10^(-10) C//kg` |
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