The figure shows two tiny 5.0-g spheres suspended from very light 1.0-m-long threads. The spheres repel each other after each one is given the same positive charge and hang at rest when θ = 4.1°. What is the charge on each sphere? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²) A) 22 nC B) 89 nC C) 360 nC D) 180 nC E) 45 nC

Question

Quizplus · Accepted Answer

The force of repulsion between the spheres can be found using Coulomb's law:

$F=\frac{kq^2}{r^2}$

where $k$ is the Coulomb constant, $q$ is the charge on each sphere, and $r$ is the distance between the spheres.

Since the spheres are at rest, the net force in the horizontal direction is zero. Therefore, the force of repulsion must be balanced by the horizontal component of the tension in the strings:

$F\sin	heta=T\sin(180^\circ-2	heta)$

where $T$ is the tension in the strings.

Substituting the expressions for $F$ and $T$ and solving for $q$, we get:

$q^2=\frac{2kmg\sin	heta}{\sin(180^\circ-2	heta)}$

$q=\sqrt{\frac{2kmg\sin	heta}{\sin(180^\circ-2	heta)}}$

Plugging in the given values, we get:

$q=\sqrt{\frac{2(9.0	imes10^9\mathrm{N\cdot m^2/C^2})(0.005\mathrm{kg})(9.81\mathrm{m/s^2})(\sin 4.1^\circ)}{\sin(180^\circ-2	imes4.1^\circ)}}$

$q\approx89\mathrm{nC}$

Therefore, the answer is B.

The Figure Shows Two Tiny 5