Question

The ratio of electric force and gravitational force between a
proton and an electron at a certain distance is
(A)10^41
B) 2.4x10^41
(C) 2.4x10^39
(D) 3.9x10^24​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{ANSWER}}}}} \\ \\ \implies \sf \: mass \: of \: ptoton \: m1 = 1.67 \times {10}^{ - 27} \: kg \\ \implies \sf \: mass \: of \: electron \: m2 = 9.1 \times {10}^{ - 31} \: kg \\ \implies \sf \: charge \: of \: proton \: q1 = 1.6 \times {10}^{ - 19} \: C \\ \implies \sf \: charge \: of \: electron \: q2 = - 1.6 \times {10}^{ - 19} \: C \\ \implies \sf \: gravitational \: constant \: G = 6.67 \times {10}^{ - 11} \: \frac{N {m}^{2} }{ {kg}^{2} } \\ \implies \sf \: coloumb \: constant \: k = 9 \times {10}^{9} \: \frac{N {m}^{2} }{ {C}^{2}} \\ \\ \star \sf \: \blue{ \bold{Formula}} \\ \\ \implies \sf \: \pink{ \bold{Fe = \frac{kq1q2}{ {r}^{2} } }} \\ \\ \implies \sf \: \green{ \bold{Fg = \frac{Gm1m2}{ {r}^{2}}}} \\ \\ \star \sf \: \blue{ \bold{Calculation}} \\ \\ \implies \sf \: { \frac{Fe}{Fg} = \frac{kq1q2}{Gm1m2} } \\ \\ \implies \sf \: \frac{Fe}{Fg} = \frac{9 \times {10}^{9} \times 1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{6.67 \times {10}^{ - 11} \times 1.67 \times {10}^{ - 27} \times 9.1 \times {10}^{ - 31} } \\ \\ \huge{\star} \: \boxed{ \bold{ \orange{ \sf{ \frac{Fe}{Fg} = 2.4 \times {10}^{41} }}}}$