Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Two particles of equal mass m and charge q are placed at a distance of 16cm. They do not experience any force. The value of $\dfrac{q}{m}$ is
A. 1
B. $\sqrt{\dfrac{\pi {{\varepsilon }_{0}}}{G}}$
C. $\sqrt{\dfrac{G}{4\pi {{\varepsilon }_{0}}}}$
D. $\sqrt{4\pi {{\varepsilon }_{0}}G}$

Answer
VerifiedVerified
465k+ views
Hint: As a first step, you could find out all the possible forces that can act on the two particles. As it is given in the question that there is no force between them, the sum of the force so found would be zero. You could accordingly assign the directions of the forces. Then, you could rearrange to get the answer.

Formula used:
Gravitational force,
${{F}_{g}}=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Coulomb’s law,
${{F}_{e}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Complete Step by step solution:
In the question, we are given two particles each of m mass and charge q separated by a distance of 16cm and don't experience any force. We are supposed to find the charge by mass ratio, that is, $\dfrac{q}{m}$
seo images

For a system such as this where two charges are kept at distance from each other, the possible forces that can act on them will be gravitational force and electrostatic force.
The gravitational force between two particles of mass m each is given by,
${{F}_{g}}=\dfrac{Gmm}{{{r}^{2}}}=\dfrac{G{{m}^{2}}}{{{r}^{2}}}$ ……………………………………………… (1)

Now, we have the electrostatic force between the two particles of identical charge q. This force could be given by using Coulomb's law.
${{F}_{e}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{r}^{2}}}$ ……………………………………………………. (2)
In the question, we are also told that the net force acting on the particles is zero. So the sum of the gravitational force and electrostatic force will be zero. That is,
$\left| {{F}_{g}} \right|=\left| {{F}_{e}} \right|$

Substituting (1) and (2),
$\Rightarrow \left| \dfrac{G{{m}^{2}}}{{{r}^{2}}} \right|=\left| \dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}^{2}}}{{{r}^{2}}} \right|$
$\Rightarrow \dfrac{{{q}^{2}}}{{{m}^{2}}}=4\pi {{\varepsilon }_{0}}G$
$\therefore \dfrac{q}{m}=\sqrt{4\pi {{\varepsilon }_{0}}G}$

Therefore, we found that that the ratio of charge to the mass is given by,
$\dfrac{q}{m}=\sqrt{4\pi {{\varepsilon }_{0}}G}$

Hence, option D is found to be the correct answer.

Note:
You may have noticed that we have taken the magnitude of the forces to be equal. This is because, as the charges are identical, the particles will obviously repel each other. Gravitational force will be attractive. So if equal magnitudes of forces are acting in opposite directions, the net force on the particles is zero.