Question

Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of $\frac{q}{m}$ is

• A. l
• B. $\sqrt{\frac{\pi {ϵ}_0}{G}}$
  
• C. $\sqrt{\frac{G}{4\pi {ϵ}_0}}$
  
• D. $\sqrt{4\pi {ϵ}_{0}G}$

Answer 1

Answer: (D)

Step 1: Given that:

The equal mass($m_1=m_2$) of the two particles= $m$

The equal charge($q_1=q_2$) of the two particles= $q$

The distance($r$) between the particles= $16cm$ = $16\times {10}^{-2}m$

Step 2: Concept and formula used:

1. Gravitational force acts between two bodies having masses.
2. The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by; $F_G=G\frac{m_1{m}_2}{r^2}$ , where G is the universal gravitational constant.
   
3. An electrostatic force acts between two electric charges.
4. The gravitational force between two charges, $q_1$ and $q_2$ separated by a distance $r$ is given by; $F_e=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$ , where ${\epsilon }_0$ is the permittivity of free space.

Step 3: Calculation of the value of $\frac{q}{m}$ :

As the particles are not experiencing any force,

It means,

The net gravitational force on the particles = The net electrostaic force on the particles.

That is,

$|\overrightarrow{F_G}|=|\overrightarrow{F_e}|$

$G\frac{m\times m}{{(16\times {10}^{-2}m)}^2}=\frac{1}{4\pi {\epsilon }_0}\frac{q\times q}{{(16\times {10}^{-2}m)}^2}$

$G\times m^2=\frac{1}{4\pi {\epsilon }_0}\times q^2$

$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\times q^2=G\times m^2$

$\Rightarrow \frac{q^2}{m^2}=4\pi {\epsilon }_{0}G$

$Takingsquarerootonbothsides,weget;$

$\Rightarrow \frac{q}{m}=\sqrt{4\pi {\epsilon }_{0}G}$

Thus,

Option D) $\sqrt{4\pi {\epsilon }_{0}G}$ is the correct option.