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Answer 1

we have given that,

force b/w the two charges is equal to weight of the ither of the charge

let us take has acceleration due to gravity be 10 m/s^2

therefore,

force is

F = mg

= 1.6 × 10^-27 kg × 10 m/s^2

= 1.6 × 10^-26 N

therefore,

by formula distance between them will be

F = k. qq/r^2

=> r^2 = k. qq/F

=> r^2 = 9 × 10^9 N- m^2/C^2 {(1.6 × 10^-19)^2 C /(1.6×10^-26)N}

=> r = 0.12 m (ans.)

Answer 2

Step-by-step explanation:

$F_e = \frac{Kq_1q_2}{ {r}^{2} } \\ m_pg = \frac{Kq_pq_p}{ {r}^{2} } \\ {r}^{2} = \frac{K {q_p}^{2} }{ m_pg} \\ r = \sqrt{\frac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{1.673 \times {10}^{ - 27} \times 10}} \\ r = \sqrt{\frac{9 \times {10}^{36} \times {(1.6 \times {10}^{ - 19} )}^{2} }{16.73 }} \\ r = 3 \times {10}^{18} \times 1.6 \times {10}^{ - 19} \times \sqrt{ \frac{1}{16.73} } \\ r = 4.8 \times {10}^{ - 1} \times 0.059 \\ r ≈ 0.028 \: m \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$