Question #109dc

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Answer 1 · 2015-11-10T18:21:02

$202.5 N$ $202.5"N"$

Coulomb's Law gives us the force $F$ $F$ between two changes $q_{2}$ $q_2$ and $q_{2}$ $q_2$ separated by a distance $r$ $r$ :

$F = \frac{1}{4 π ε_{0}} . \frac{q_{1} \times q_{2}}{r^{2}}$ $F=1/(4piepsilon_0).(q_1xxq_2)/(r^2)$

This simplifies to:

$F = k . \frac{q_{1} \times q_{2}}{r^{2}}$ $F=k.(q_1xxq_2)/(r^2)$

Where $k = 9 \times 10^{9} m/F$ $k=9xx10^(9)"m/F"$

This distance between the two spheres is taken as the distance between their two centres. So $r = 2 \times 0.2 = 0.4 m$ $r=2xx0.2=0.4"m"$

$:.F=(9xx10^(9)xx(60xx10^(-6))^(2))/(0.4^2)$

$:.F=202.5"N"$