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 q/m is

Answer 1

Final Answer :
(G/k)^1/2 where G is gravitational constant and K is Coulomb Constant.

Steps:
1) Since, particle is in horizontal frame, so there will not be any effect from Gravitation force due to earthin Horizontal direction

2) But, there will be attractive force due to Attraction between masses by Gravitational force due to other mass m.
And There will be repulsive force due to like charges repel each other.
3) Since, system is in equilibrium.
Therefore. Net Force along Horizontal direction us 0 .

Answer 2

Answer:

$\frac{\mathrm{q}}{\mathrm{m}}=\sqrt{4 \pi \epsilon_{0} \mathrm{G}}$

Explanation:

Solution

Step 1: Electrostatic force between particles

Applying Coulomb's law,

$\mathrm{F}_{\mathrm{e}}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}^{2}}{(0.16 \mathrm{~m})^{2}}$

Step 2: Gravitational force between particles

Applying Newton's law of universal gravitation,

$F_{G}=G \frac{m_{1} m_{2}}{r^{2}}=G \frac{m^{2}}{(0.16)^{2}}$

Step 3: Net force experienced

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

$\therefore \mathrm{F}_{\mathrm{e}}=\mathrm{F}_{\mathrm{G}}$

From (1) and (2)

$\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}^{2}}{(0.16)^{2}}=\mathrm{G} \frac{\mathrm{m}^{2}}{(0.16)^{2}}$

$\Rightarrow \frac{\mathrm{q}}{\mathrm{m}}=\sqrt{4 \pi \epsilon_{0} \mathrm{G}}$

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