Q. Assume electrons of $10^{23}$ carbon atoms are put at the south pole of the earth and the nuclei at the north pole of the earth (radius $=6400km$ ). Find the force between the charges.
NTA AbhyasNTA Abhyas 2020
Solution:
Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
According to coulumb's law,
$F=\frac{k q_{1} q_{2}}{r^{2}}$
Number of protons and elctrons in Carbon atom is $6$ each.
hence,
$\Rightarrow F_{n e t}=\left(\frac{k q^{2}}{r^{2}}\right)$
$\Rightarrow F_{n e t}=8.98\times \left(10\right)^{9}\times \frac{\left(6 \times \left(10\right)^{23} \times 1 . 6 \times \left(10\right)^{- 19}\right)^{2}}{\left(2 \times 6400 \times 1000\right)^{2}}=\frac{8 . 98 \times 36 \times 2 . 56 \times \left(10\right)^{17}}{4 \times 4096 \times \left(10\right)^{10}}=0.050\times \left(10\right)^{7}=5\times \left(10\right)^{5}$
According to coulumb's law,
$F=\frac{k q_{1} q_{2}}{r^{2}}$
Number of protons and elctrons in Carbon atom is $6$ each.
hence,
$\Rightarrow F_{n e t}=\left(\frac{k q^{2}}{r^{2}}\right)$
$\Rightarrow F_{n e t}=8.98\times \left(10\right)^{9}\times \frac{\left(6 \times \left(10\right)^{23} \times 1 . 6 \times \left(10\right)^{- 19}\right)^{2}}{\left(2 \times 6400 \times 1000\right)^{2}}=\frac{8 . 98 \times 36 \times 2 . 56 \times \left(10\right)^{17}}{4 \times 4096 \times \left(10\right)^{10}}=0.050\times \left(10\right)^{7}=5\times \left(10\right)^{5}$