Question

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? 21.27 ** cp an electron is projected with an initial speed $v_0 = 1.60\times10^{6} m/s$ into the uniform field between two parallel plates (fig. e21.27). assume that the field between the plates is uniform and directed vertically downward and that the field outside the plates is zero. the electron enters the field at a point midway between the plates. (a) if the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. (b) suppose that the electron in fig. e21.27 is replaced by a proton with the same initial speed $v_0$. would the proton hit one of the plates? if not, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (c) compare the paths traveled by the electron and the proton, and explain the differences. (d) discuss whether it is reasonable to ignore the effects of gravity for each particle.

Explanation:

Step1: Recall Coulomb's law

$F = k\frac{q_1q_2}{r^2}$, where $F$ is the force, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are charges, and $r$ is the distance between charges. Given $r=15.0\ cm = 0.15\ m$ and $F = 0.220\ N$.

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

Substitute into Coulomb's law: $F=k\frac{q\cdot q}{r^2}=k\frac{q^{2}}{r^{2}}$. Then $q^{2}=\frac{F\cdot r^{2}}{k}$. So $q=\sqrt{\frac{F\cdot r^{2}}{k}}=\sqrt{\frac{0.220\times(0.15)^{2}}{9\times 10^{9}}}=\sqrt{\frac{0.220\times0.0225}{9\times 10^{9}}}=\sqrt{\frac{0.00495}{9\times 10^{9}}}=\sqrt{5.5\times 10^{-13}}\approx7.42\times 10^{-7}\ C$.

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

Substitute into Coulomb's law: $F=k\frac{q\times4q}{r^{2}}=k\frac{4q^{2}}{r^{2}}$. Then $q^{2}=\frac{F\cdot r^{2}}{4k}$. So $q=\sqrt{\frac{F\cdot r^{2}}{4k}}=\sqrt{\frac{0.220\times(0.15)^{2}}{4\times9\times 10^{9}}}=\sqrt{\frac{0.220\times0.0225}{36\times 10^{9}}}=\sqrt{\frac{0.00495}{36\times 10^{9}}}=\sqrt{1.375\times 10^{-13}}\approx3.71\times 10^{-7}\ C$. One charge is $q\approx3.71\times 10^{-7}\ C$ and the other is $4q\approx1.48\times 10^{-6}\ C$.

Answer:

(a) $q\approx7.42\times 10^{-7}\ C$
(b) $q_1\approx3.71\times 10^{-7}\ C$, $q_2\approx1.48\times 10^{-6}\ C$