Two electrons are $20.0 \mathrm {~mm}$ apart at closest approach. What is the magnitude of the maximum electric force that they exert on each other? $\left( e = 1.60 \times 10 ^ { - 19 } \mathrm { C } , k = 1 / 4 \pi \varepsilon _ { 0 } = 9.010 ^ { 9 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { C } ^ { 2 } \right)$

Question

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The electric force between two charges is given by Coulomb's law: $F = k \frac{|q_1 q_2|}{r^2}$, where $k = 9.0 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2/\mathrm{C}^2$, $q_1 = q_2 = 1.60 \times 10^{-19} \mathrm{~C}$, and $r = 20.0 \mathrm{~mm} = 0.020 \mathrm{~m}$. Substituting these values, $F = 9.0 \times 10^9 \frac{(1.60 \times 10^{-19})^2}{(0.020)^2} = 5.8 \times 10^{-25} \mathrm{~N}$.

Two Electrons Are 20.0 mm20.0 \mathrm {~mm}20.0 mm Apart at Closest Approach A) 5.8×10−27 N5.8 \times 10 ^ { - 27 } \mathrm {~N}5.8×10−27 N

Two Electrons Are $20.0 \mathrm {~mm}$ Apart at Closest Approach A) $5.8 \times 10 ^ { - 27 } \mathrm {~N}$