Question 40: Electrostats & 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:

$\$ $(	ext{Take } k = \dfrac{1}{4\pi \varepsilon_0} 	ext{ and assume } kQ^2 > Gm^2 	ext{ where } G 	ext{ is the Gravitational constant})$

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,
$K_i = \dfrac{1}{2}mu^2 + \dfrac{1}{2}mu^2 = mu^2$ $\$

Initial electrostatic potential energy is,
$U_{ei} = \dfrac{kQ^2}{4R}$ $\$

Initial gravitational potential energy is,
$U_{gi} = -\dfrac{Gm^2}{4R}$ $\$

When the spheres just touch, their centres are separated by a distance equal to the sum of their radii $(R + R = 2R)$. To find the minimum $u$, we assume they come to a momentary rest at this point. $\$

Total final kinetic energy, $K_f = 0$
$\$
Final electrostatic potential energy is,
$U_{ef} = \dfrac{kQ^2}{2R}$
$\$
Final gravitational potential energy is,
$U_{gf} = -\dfrac{Gm^2}{2R}$
$\$
The total energy remains constant:
$K_i + U_{ei} + U_{gi} = K_f + U_{ef} + U_{gf}$
$\$
$\Rightarrow mu^2 + \dfrac{kQ^2}{4R} - \dfrac{Gm^2}{4R} = 0 + \dfrac{kQ^2}{2R} - \dfrac{Gm^2}{2R}$
$\$
$\Rightarrow mu^2 = \left(\dfrac{kQ^2}{2R} - \dfrac{kQ^2}{4R}ight) - \left(\dfrac{Gm^2}{2R} - \dfrac{Gm^2}{4R}ight)$
$\$
$\Rightarrow mu^2 = \dfrac{kQ^2}{4R} - \dfrac{Gm^2}{4R}$
$\$
$\Rightarrow mu^2 = \dfrac{kQ^2}{4R}\left(1 - \dfrac{Gm^2}{kQ^2}ight)$
$\$
$\Rightarrow u^2 = \dfrac{kQ^2}{4mR}\left(1 - \dfrac{Gm^2}{kQ^2}ight)$
$\$
$\Rightarrow u = \sqrt{\dfrac{kQ^2}{4mR}\left(1 - \dfrac{Gm^2}{kQ^2}ight)}$

Answer

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

Answer

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

Answer

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

Answer

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

Electrostats & Capacitance

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