# Question #88660

Nov 3, 2017

There will be a force of replusion, ${F}_{r}$, because both charges have the same sign. Coulomb's Law, in equation form, tells us that the repulsive force will be
${F}_{r} = \frac{k \cdot {q}_{1} \cdot {q}_{2}}{r} ^ 2$

Plugging in the data and working it out gives us

${F}_{r} = \frac{9.0 \times {10}^{9} N \cdot {m}^{2} / {C}^{2} \cdot \left(- 22.0 \times {10}^{-} 9 C\right) \cdot \left(- 22.0 \times {10}^{-} 9 C\right)}{0.06 m} ^ 2$

${F}_{r} = \frac{9.0 \times {10}^{9} N \cdot {m}^{2} / \cancel{{C}^{2}} \cdot \left(- 22.0 \times {10}^{-} 9 \cancel{C}\right) \cdot \left(- 22.0 \times {10}^{-} 9 \cancel{C}\right)}{0.0036 {m}^{2}}$

${F}_{r} = \frac{9.0 \times {10}^{9} N \cdot \cancel{{m}^{2}} \cdot - 22.0 \times - 22 \times {.010}^{-} 18}{0.0036 \cancel{{m}^{2}}}$

${F}_{r} = \frac{9.0 \times {10}^{9} N \cdot 22.0 \times 22.0 \times {.010}^{-} 18}{0.0036}$

${F}_{r} = \frac{9.0 N \cdot 22.0 \times 22.0 \times {10}^{-} 9}{0.0036} = 1 , 210 , 000 N$

${F}_{r} = 1 , 210 , 000 N = 1.21 \times {10}^{-} 3 N$

I hope this helps,
Steve