Fg and Fg represent the gravitational and electric forces between the electrons at distance 10 cm, then the ratio of (Fg/Fe) will be

Question

Fg and Fg represent the gravitational and electric forces between the electrons at distance 10 cm, then the ratio of (Fg/Fe) will be (A)  1042  (B) 10 (C) 1 (D)  10-43

Tardigrade Admin · Accepted Answer

(Fg/Fe)=(G(m.m/r2)/(1)4π ε 0.(q1q2)r2=(6.67× 10-11× (9.1× 10-31)/9× 109× (1.6× 10-19)2) ≈ /10-43)

Q. $F_{g}$ and $F_{g}$ represent the gravitational and electric forces between the electrons at distance $10 c m$ , then the ratio of $\frac{F _{g}}{F _{e}}$ will be

$10^{42}$

10

1

$10^{- 43}$

$\frac{F _{g}}{F _{e}} = \frac{G \frac{m . m}{r ^{2}}}{\frac{1}{4 π ε _{0}} . \frac{q _{1} q _{2}}{r ^{2}}} = \frac{6.67 \times 10 ^{- 11} \times ( 9.1 \times 10 ^{- 31} )}{9 \times 10 ^{9} \times ( 1.6 \times 10 ^{- 19} ) ^{2}}$ $\approx 10^{- 43}$

Q. Fg​ and Fg​ represent the gravitational and electric forces between the electrons at distance 10cm, then the ratio of Fe​Fg​​ will be

1042

10

1

10−43

Fe​Fg​​=4πε0​1​.r2q1​q2​​Gr2m.m​​=9×109×(1.6×10−19)26.67×10−11×(9.1×10−31)​ ≈10−43

Q. $F_{g}$ and $F_{g}$ represent the gravitational and electric forces between the electrons at distance $10 c m$ , then the ratio of $\frac{F _{g}}{F _{e}}$ will be

$10^{42}$

$10^{- 43}$

$\frac{F _{g}}{F _{e}} = \frac{G \frac{m . m}{r ^{2}}}{\frac{1}{4 π ε _{0}} . \frac{q _{1} q _{2}}{r ^{2}}} = \frac{6.67 \times 10 ^{- 11} \times ( 9.1 \times 10 ^{- 31} )}{9 \times 10 ^{9} \times ( 1.6 \times 10 ^{- 19} ) ^{2}}$ $\approx 10^{- 43}$