## College Physics (4th Edition)

The charge on one sphere is $1.29\times 10^{-6}~C$ The charge on the other sphere is $6.21\times 10^{-6}~C$
Let $q$ be the charge on one sphere. Then the charge on the other sphere is $7.50\times 10^{-6}~C - q$ We can find the charge $q$: $F = \frac{k~\vert q_1 \vert~\vert q_2 \vert}{r^2}$ $20.0~N = \frac{(9.0\times 10^9~N~m^2/C^2)(q)(7.50\times 10^{-6}~C-q)}{(0.060~m)^2}$ $0.072~N~m^2 = (67,500~N~m^2/C)~q - (9.0\times 10^9~N~m^2/C^2)~q^2$ $(9.0\times 10^9~N~m^2/C^2)~q^2 - (67,500~N~m^2/C)~q + 0.072~N~m^2 = 0$ We can use the quadratic formula to find $q$: $q = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $q = \frac{67,500 \pm\sqrt{(-67,500)^2 - (4)(9.0\times 10^9)(0.072)}}{(2)(9.0\times 10^9)}$ $q = 1.29\times 10^{-6}~C, 6.21\times 10^{-6}~C$ The quadratic formula gives us both possible values of $q$. The charge on one sphere is $1.29\times 10^{-6}~C$ The charge on the other sphere is $6.21\times 10^{-6}~C$