Two particles of equal mass m and charge q ar | vedclass

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(D) For the particles to experience no net force,the gravitational force of attraction must be balanced by the electrostatic force of repulsion.$|\vec

Vedclass · Accepted Answer

$\sqrt{4\pi \varepsilon_0 G}$

Vedclass · Answer

(D) For the particles to experience no net force,the gravitational force of attraction must be balanced by the electrostatic force of repulsion.$|\vec{F_G}| = |\vec{F_e}|$$G \frac{m^2}{r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q^2}{r^2}$Canceling $r^2$ from both sides,we get:$G m^2 = \frac{1}{4\pi \varepsilon_0} q^2$Rearranging the terms to find $\frac{q}{m}$:$\frac{q^2}{m^2} = 4\pi \varepsilon_0 G$$\frac{q}{m} = \sqrt{4\pi \varepsilon_0 G}$

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

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