Question

A particle A having a charge of 2 x 10^-6 and a mass of 100g is placed at the bottom of a smooth inclined plane of inclination 30 degree. Where should another charge B, having same charge and mass, be placed on the incline ao that it may remain in equilibrium?
(Please explain it thoroughly)

Answer 1

A charged particle charge Q = 2 × 10^-6C and mass m = 100g is placed at bottom of a smooth inclined plane another charged particle of same charge and mass are placed r distance from first particle as shown in figure.

Now, at equilibrium
mgsin30° = Coulombs force = Kq²/r²
100 × 10⁻³ × 10 × 1/2 = 9 × 10⁹ × (2 × 10⁻⁶)²/r²
1/2 = 9 × 10⁹ × 4 × 10⁻¹²/r²
r² = 72 × 10⁻³
r² = 0.072
taking square root both sides,
r = 0.268 m or, 26.8 cm

Hence, charge B should placed at 26.8 cm from charge A

Answer 2

Answer:

The another charge should be kept 27 cm.

Explanation:

Given that,

Charge $q = 2\times10^{-6}$

Mass m = 100 g

Angle = 30°

The force along the inclination on the particle due to gravity

$F =mg\sin\theta$

$F =0.1\times9.8\times\sin30^{\circ}$

$F = 0.1\times9.8\times0.5$

$F = 0.49\ N$

The charges are same on the particle so the charges will be repel to each other.

Let the another charge is kept at r meter away

At equilibrium,

The force along the inclination = coulomb force

$F = \dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}r^2}$

$r^2=\dfrac{q_{1}q_{2}}{4\pi\epsilon_{0}F}$

$r^2=\dfrac{2\times10^{-6}\times2\times10^{-6}\times9\times10^{9}}{0.49}$

$r=\sqrt{0.073}$

$r = 0.27\ m$

$r = 27\ cm$

Hence, The another charge should be kept 27 cm away.