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  4. The electrostatics force of repulsion between two positively charged ions carrying equal charge is 3.7× 10-7N, when they are separated by a distance of 5 overset texto mathop textA . What are the number of electrons are missing each ion?

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Q. The electrostatics force of repulsion between two positively charged ions carrying equal charge is $ 3.7\times {{10}^{-7}}N, $ when they are separated by a distance of 5 $ \overset{\text{o}}{\mathop{\text{A}}}\, $ . What are the number of electrons are missing each ion?

CMC MedicalCMC Medical 2015

Solution:

From Coulambs law $ F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{{{q}_{1}}\,{{q}_{2}}}{{{r}^{2}}} $ $ 3.7\times {{10}^{-9}}=9\times {{10}^{9}}\times \frac{{{q}^{2}}}{{{(5\times {{10}^{-10}})}^{2}}} $ $ q=3.2\times {{10}^{-19}}c $ $ q=nc $ $ n=\frac{q}{c}=\frac{3.2\times {{10}^{-19}}}{1.6\times {{10}^{-10}}}=2 $