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Question: A positive ion $A$ and a negative ion $B$ has charges $6.67 \times 10^{-19}C$ and $9.6 \times 10^{-1...

A positive ion AA and a negative ion BB has charges 6.67×10−19C6.67 \times 10^{-19}C and 9.6×10−10C9.6 \times 10^{-10}C, and masses 19.2×10−2719.2 \times 10^{-27}kg and 9×10−279 \times 10^{-27}kg respectively. At an instant, the ions are separated by a certain distance rr. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is P×10−13P \times 10^{-13}, where the value of PP is _________.

(Take 14πϵ0=9×109Nm2C−1\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 Nm^2C^{-1} and universal gravitational constant as 6.67×10−11Nm2kg−26.67 \times 10^{-11} Nm^2 kg^{-2})

Answer

5.0 × 10^57

Explanation

Solution

The ratio of the magnitudes of the electrostatic force to the gravitational force is given by:

FeFg=14πϵ0∣qAqB∣r2GmAmBr2=14πϵ01G∣qAqB∣mAmB\frac{F_e}{F_g}=\frac{\frac{1}{4\pi\epsilon_0}\frac{|q_Aq_B|}{r^2}}{G\frac{m_Am_B}{r^2}}=\frac{1}{4\pi\epsilon_0}\frac{1}{G}\frac{|q_Aq_B|}{m_Am_B}.

Given:

  • 14πϵ0=9.0×109 Nm2C−1\frac{1}{4\pi\epsilon_0}=9.0\times10^9\,\mathrm{Nm^2C^{-1}}
  • G=6.67×10−11 Nm2kg−2G=6.67\times10^{-11}\,\mathrm{Nm^2kg^{-2}}
  • qA=6.67×10−19 Cq_A=6.67\times10^{-19}\,\mathrm{C}
  • qB=9.6×10−10 Cq_B=9.6\times10^{-10}\,\mathrm{C}
  • mA=19.2×10−27 kgm_A=19.2\times10^{-27}\,\mathrm{kg}
  • mB=9.0×10−27 kgm_B=9.0\times10^{-27}\,\mathrm{kg}

Step 1: Calculate the product of charges:

qAqB=6.67×10−19×9.6×10−10=6.67×9.6×10−29≈64.032×10−29=6.4032×10−28q_Aq_B=6.67\times10^{-19}\times9.6\times10^{-10}=6.67\times9.6\times10^{-29}\approx64.032\times10^{-29}=6.4032\times10^{-28}.

Step 2: Calculate the product of masses:

mAmB=19.2×10−27×9.0×10−27=172.8×10−54=1.728×10−52m_Am_B=19.2\times10^{-27}\times9.0\times10^{-27}=172.8\times10^{-54}=1.728\times10^{-52}.

Step 3: Compute the ratio:

FeFg=(9.0×1096.67×10−11)(6.4032×10−281.728×10−52)\frac{F_e}{F_g}=\left(\frac{9.0\times10^9}{6.67\times10^{-11}}\right)\left(\frac{6.4032\times10^{-28}}{1.728\times10^{-52}}\right).

First, compute the prefactor:

9.0×1096.67×10−11≈9.06.67×1020≈1.35×1020\frac{9.0\times10^9}{6.67\times10^{-11}}\approx\frac{9.0}{6.67}\times10^{20}\approx1.35\times10^{20}.

Next, compute the fractional part:

6.4032×10−281.728×10−52=6.40321.728×1024≈3.705×1024\frac{6.4032\times10^{-28}}{1.728\times10^{-52}}=\frac{6.4032}{1.728}\times10^{24}\approx3.705\times10^{24}.

Multiply them:

FeFg≈1.35×1020×3.705×1024≈5.0×1044\frac{F_e}{F_g}\approx1.35\times10^{20}\times3.705\times10^{24}\approx5.0\times10^{44}.

The problem states that the ratio is expressed as P×10−13P\times10^{-13}. So, we write:

5.0×1044=P×10−135.0\times10^{44}=P\times10^{-13}.

Thus,

P=5.0×1044×1013=5.0×1057P=5.0\times10^{44}\times10^{13}=5.0\times10^{57}.