Question

Two charged particles each having a charge '\(+q\)' and mass 'm' are kept at a distance 'd'. If they are in equilibrium under the gravitational force and the electric force between them, then the ratio \(\frac{q}{m}\) (specific charge) of each particle is (\(G\)- Universal gravitation constant, \(\varepsilon _{0}\) = permittivity of free space)

• A. \(G/4\pi \varepsilon _{0}\)
• B. \(4\pi \varepsilon _{0}G\)
• C. \(\sqrt{4\pi \varepsilon _{0}G}\)
• D. \(\sqrt{\frac{4\pi \varepsilon _{0}}{G}}\)

Hint:

Equate the Coulomb repulsion to the gravitational attraction between the two identical particles.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The particles repel electrically and attract gravitationally. In equilibrium the two forces are equal in magnitude.

Step 2: Equate the Forces
\(\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^{2}}{d^{2}}=\dfrac{Gm^{2}}{d^{2}}\)

Step 3: Calculation
\(\dfrac{q^{2}}{m^{2}}=4\pi\varepsilon_0G\)
\(\dfrac{q}{m}=\sqrt{4\pi\varepsilon_0G}\)

Step 4: Compare the options
This is the third option.

Final Answer:
The specific charge is the square root of 4 pi epsilon naught G. \[ \boxed{\text{(C) }\sqrt{4\pi\varepsilon_{0}G}} \]