Question

Two identical charged particles of same mass are placed at such a distance that the coulomb's force of repulsion is equal to the
gravitational force of attraction. Then the specific charge of these particles is:
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Answer 1

Answer:

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Answer 2

Answer:

The specific charge of the particle is $\sqrt{\frac{G}{K}}$.

Explanation:  

Given:

Two identical charged particles of the same mass are placed.

As we all know, the Coulomb's constant is denoted by K and the Gravitational constant is denoted by G.

Also,

Let $q_{1}$ be the charge of the first particle.

Let $q_{2}$ be the charge of the second particle.

Let $m_{1}$ be the mass of the first particle.

Let $m_{2}$ be the mass of the second particle.

The distance between two particles is denoted by r.

Here,

$m_{1} =m_{2} \\q_{1} =q_{2}$                             ∴ Because the particles are identical.

Then,

According to the question,

Coulomb's force = Gravitation force

$\frac{Kq_{1} q_{2} }{r^{2} } =\frac{Gm_{1} m_{2} }{r^{2} } \\\frac{Kq\times q }{r^{2} } =\frac{Gm\times m }{r^{2} } \\\frac{q}{m}=\sqrt{\frac{G}{K}}$

Now,

By the equation,

Specific charge =$\frac{q}{m}$

Specific charge $=\sqrt{\frac{G}{K}}$                                    ∴ $\frac{q}{m}=\sqrt{\frac{G}{K}}$

So, the specific charge is equal to $\sqrt{\frac{G}{K}}$.