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

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Answer 1

$Step 1: Electrostatic force between particles$

Applying Coulomb's law,
$F_{e} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}} = \frac{1}{4 π ϵ _{0}} \frac{q ^{2}}{( 0 . 1 6 m ) ^{2}}$ $. . . . (1)$

$Step 2: Gravitational force between particles$
Applying Newton's law of universal gravitation,
$F_{G} = G \frac{m _{1} m _{2}}{r ^{2}} = G \frac{m ^{2}}{( 0 . 1 6 ) ^{2}}$ $. . . . (2)$

$Step 3: Net force experienced$
Since net force experienced by the particle is zero, therefore electrostatic force and gravitational force would be equal.
$∴ F_{e} = F_{G}$
From $(1)$ and $(2)$

$\frac{1}{4 π ϵ _{0}} \frac{q ^{2}}{( 0 . 1 6 ) ^{2}} = G \frac{m ^{2}}{( 0 . 1 6 ) ^{2}}$

$\Rightarrow \frac{q}{m} = 4 π ϵ_{0} G$

Hence Option D is correct

Answer 2

Step 1: Electrostatic force between particles

Applying Coulomb's law,

F_{e} = \frac{1}{4 π ϵ _{0}} \frac{q _{1} q _{2}}{r ^{2}} = \frac{1}{4 π ϵ _{0}} \frac{q ^{2}}{( 0 . 1 6 m ) ^{2}}

. . . . (1)

Step 2: Gravitational force between particles

Applying Newton's law of universal gravitation,

F_{G} = G \frac{m _{1} m _{2}}{r ^{2}} = G \frac{m ^{2}}{( 0 . 1 6 ) ^{2}}

. . . . (2)

Step 3: Net force experienced

Since net force experienced by the particle is zero, therefore electrostatic force and gravitational force would be equal.

∴ F_{e} = F_{G}

From

(1)

and

(2)

\frac{1}{4 π ϵ _{0}} \frac{q ^{2}}{( 0 . 1 6 ) ^{2}} = G \frac{m ^{2}}{( 0 . 1 6 ) ^{2}}

\Rightarrow \frac{q}{m} = 4 π ϵ_{0} G

Hence Option D is correct