Question 178: Electrostatics & Capacitance

Question

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$. Their centers have an initial separation of $4R$. Both the spheres are given an initial speed of $u$ towards each other. The minimum value of $u$, so that they can just touch each other is:
(Take $k = \frac{1}{4 \pi \epsilon_0}$ and assume $kQ^2 > Gm^2$ where $G$ is the Gravitational constant)

Accepted Answer

At the start, the spheres are moving towards each other.
Since there are two spheres of mass $m$ moving with speed $u$, so the initial kinetic energy is,
$$ \mathrm{K}_{\mathrm{i}}=\frac{1}{2} \mathrm{mu}^2+\frac{1}{2} \mathrm{mu}^2=\mathrm{mu}^2 $$
Initial electrostatic potential energy is,
$$ \mathrm{U}_{\mathrm{ei}}=\frac{\mathrm{kQ}^2}{4 \mathrm{R}} $$
Initial gravitational potential energy is :
$$ U_{g i}=-\frac{G \mathrm{~m}^2}{4 \mathrm{R}} $$
When the spheres just touch, their centres are separated by a distance equal to the sum of their radii $(R+R=2 R)$. To find the minimum $u$, we assume they come to a momentary rest at this point.
Total final kinetic energy, $\mathrm{K}_{\mathrm{f}}=0$
Final electrostatic potential energy is,
$$ \mathrm{U}_{\mathrm{ef}}=\frac{\mathrm{k} \mathrm{Q}^2}{2 \mathrm{R}} $$
Final gravitational potential energy is :
$$ \mathrm{U}_{\mathrm{gf}}=-\frac{\mathrm{Gm}^2}{2 \mathrm{R}} $$
The total energy remains constant :
$$ K_i+U_{e i}+U_{g i}=K_f+U_{e f}+U_{g f} $$
$$\Rightarrow $$ $\mathrm{mu}^2+\frac{\mathrm{kQ}^2}{4 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}=0+\frac{\mathrm{kQ}^2}{2 \mathrm{R}}-\frac{\mathrm{Gm}^2}{2 \mathrm{R}}$
$$\Rightarrow $$ $\mathrm{mu}^2=\left(\frac{\mathrm{kQ}^2}{2 \mathrm{R}}-\frac{\mathrm{kQ}^2}{4 \mathrm{R}}\right)-\left(\frac{\mathrm{Gm}^2}{2 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}\right)$
$$\Rightarrow $$ $\mathrm{mu}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{R}}-\frac{\mathrm{Gm}^2}{4 \mathrm{R}}$
$$\Rightarrow $$ $\mathrm{mu}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{R}}\left(1-\frac{\mathrm{Gm}^2}{\mathrm{kQ}^2}\right)$
$$\Rightarrow $$ $\mathrm{u}^2=\frac{\mathrm{kQ}^2}{4 \mathrm{mR}}\left(1-\frac{\mathrm{Gm}^2}{\mathrm{kQ}^2}\right)$
$$\Rightarrow $$ $u=\sqrt{\frac{k Q^2}{4 m R}\left(1-\frac{G m^2}{k Q^2}\right)}$

Answer

$ \sqrt{\frac{kQ^2}{4mR} \left(1 + \frac{Gm^2}{kQ^2} \right)} $

Answer

$ \sqrt{\frac{kQ^2}{2mR} \left(1 - \frac{Gm^2}{2kQ^2} \right)} $

Answer

$ \sqrt{\frac{kQ^2}{2mR} \left(1 - \frac{Gm^2}{kQ^2} \right)} $

Answer

$ \sqrt{\frac{kQ^2}{4mR} \left(1 - \frac{Gm^2}{kQ^2} \right)} $

Electrostatics & Capacitance

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