Question

phys 2205 - homework - chapter 21
21.9 two small plastic spheres are given positive electric charges. when they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 n. what is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Explanation:

Step1: Recall Coulomb's law

$F = k\frac{q_1q_2}{r^{2}}$, where $F$ is the force between the charges, $k = 9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. First, convert $r = 15.0\ cm=0.15\ m$.

Step2: Solve part (a) when $q_1 = q_2=q$

Substitute into Coulomb's law: $0.220\ N=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{q\times q}{(0.15\ m)^{2}}$. Rearrange for $q^{2}$: $q^{2}=\frac{0.220\ N\times(0.15\ m)^{2}}{9\times 10^{9}\ N\cdot m^{2}/C^{2}}$. Then $q^{2}=\frac{0.220\times0.0225}{9\times 10^{9}}\ C^{2}=5.5\times 10^{-13}\ C^{2}$. So $q=\sqrt{5.5\times 10^{-13}\ C^{2}}\approx7.42\times 10^{-7}\ C$.

Step3: Solve part (b) when $q_1 = 4q_2$

Let $q_2 = q$ and $q_1 = 4q$. Substitute into Coulomb's law: $0.220\ N=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{4q\times q}{(0.15\ m)^{2}}$. Rearrange for $q^{2}$: $q^{2}=\frac{0.220\ N\times(0.15\ m)^{2}}{4\times9\times 10^{9}\ N\cdot m^{2}/C^{2}}$. Then $q^{2}=1.375\times 10^{-13}\ C^{2}$. So $q=\sqrt{1.375\times 10^{-13}\ C^{2}}\approx3.71\times 10^{-7}\ C$. The charges are $q_1 = 4q\approx1.48\times 10^{-6}\ C$ and $q_2 = q\approx3.71\times 10^{-7}\ C$.

Answer:

(a) $7.42\times 10^{-7}\ C$
(b) $1.48\times 10^{-6}\ C$ and $3.71\times 10^{-7}\ C$