Suppose that $1.0 g$ of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on earth? (Given that the radius of earth is $6400 k m$ )

510 k N

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460 k N

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710 k N

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220 k N

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Solution

The correct option is A $510 k N$
Number of electrons or protons in 1.00 g of hydrogen is
$n = 2 [\frac{1}{2} \times 6.023 \times 10^{23}] = 6.023 \times 10^{23}$
Magnitude of charge on north or south pole is
$q = n e = 6.23 \times 10^{23} \times 1.6 \times 10^{- 19} = 96368 C$
$F = \frac{k q^{2}}{r^{2}} = \frac{9 \times 10^{9} \times (96368)^{2}}{(2 \times 6400 \times 10^{3})^{2}} = 5.10 \times 10^{5} N$
$= 510 k N$

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Suppose that 1.0 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on earth? (Given that the radius of earth is 6400 km)

The correct option is A 510 kNNumber of electrons or protons in 1.00 g of hydrogen isn=2[12×6.023×1023]=6.023×1023Magnitude of charge on north or south pole isq=ne=6.23×1023×1.6×10−19=96368CF=kq2r2=9×109×(96368)2(2×6400×103)2=5.10×105N=510kN

Suppose that $1.0 g$ of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on earth? (Given that the radius of earth is $6400 k m$ )