Consider two identical metallic spheres of radius each having charge and mass . Their centers have an initial separation of

Question

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$ . Their centers have an initial separation of $4 R$ . 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

$(Take k = \frac{1}{4 π ε_{0}} and assume k Q^{2} > G m^{2} where G is the Gravitational constant)$ [2026]

Smart Achievers · Accepted Answer

(1)

$Using energy conservation$

$m$

$\Rightarrow m u^{2} = \frac{K Q^{2}}{4 R} - \frac{G m^{2}}{4 R}$

$\Rightarrow u = \sqrt{\frac{K Q^{2}}{4 m R} (1 - \frac{G m^{2}}{K Q^{2}})}$

Solution

1 $\sqrt{\frac{k Q^{2}}{4 m R} (1 - \frac{G m^{2}}{k Q^{2}})}$

2 $\sqrt{\frac{k Q^{2}}{4 m R} (1 + \frac{G m^{2}}{k Q^{2}})}$

3 $\sqrt{\frac{k Q^{2}}{2 m R} (1 - \frac{G m^{2}}{k Q^{2}})}$

4 $\sqrt{\frac{k Q^{2}}{2 m R} (1 - \frac{G m^{2}}{2 k Q^{2}})}$

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Solution

1 kQ24mR(1-Gm2kQ2)

2 kQ24mR(1+Gm2kQ2)

3 kQ22mR(1-Gm2kQ2)

4 kQ22mR(1-Gm22kQ2)

Ans. (1) Using energy conservation KQ24R-Gm24R+12mu2×2=KQ22R-Gm22R ⇒mu2=KQ24R-Gm24R ⇒u=KQ24mR(1-Gm2KQ2)

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1 $\sqrt{\frac{k Q^{2}}{4 m R} (1 - \frac{G m^{2}}{k Q^{2}})}$

2 $\sqrt{\frac{k Q^{2}}{4 m R} (1 + \frac{G m^{2}}{k Q^{2}})}$

3 $\sqrt{\frac{k Q^{2}}{2 m R} (1 - \frac{G m^{2}}{k Q^{2}})}$

4 $\sqrt{\frac{k Q^{2}}{2 m R} (1 - \frac{G m^{2}}{2 k Q^{2}})}$