Question

A particle A with a charge of 2.0 × 10−6 C and a mass of 100 g is placed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium?

Answer 1

Charge B should placed at 27m from the bottom.

Charge = Q = 2 × 10^-6C (Given)

Mass = m = 100g  (Given)

Inclination plane = 30°  (Given)

According to the Coulombs force = Kq²/r²

At equilibrium -

Coulombs force = Kq²/r² = mgsinФ

= 9 × 10`9 × 4 × 10`12 / r² = 9.8 × m × 1/2.

r² = 18 × 4 × 10³ / m × 9.8

= 72 × 10³ / 9.8 × 10`-1

= 7.34 × 10`-1

r = 2.709 × 10`-1

Therefore, charge B should placed at 27m from the bottom.

Answer 2

At distance of $2.70924 \times 10^{-1} metre$ another particle B, with the same charge and mass, be placed on the incline so that it may remain in equilibrium.

Explanation:

$\begin{equation}\text { Charging particle } \mathrm{Q}=2 \times 10^{-6} \mathrm{C}\end$

A particle mass m = 100g

Another charged particle of the same charge is placed at the bottom of a smooth inclined plane and the mass is placed r away from the first particle.

Now,                              

We know that.  

$\begin{equation}m g \sin 30^{\circ}=\text { Coulombs force }=\frac{K q^{2}}{r^{2}}\end$

For equillibrium,

Here, $q_1$ = $q_2$ = $q^2$

$\begin{equation}\frac{\mathrm{k} \mathrm{q}{^2}}{\mathrm{r}^{2}}=\mathrm{mg} \sin \theta\end$

$\begin{equation}\frac{9 \times 10^{9} \times 4 \times 10^{-12}}{r^{2}}=m \times 9.8 \times \frac{1}{2}\end$

$\begin{equation}r^{2}=\frac{18 \times 4 \times 10^{-3}}{m \times 9.8}\end$

    $\begin{equation}=\frac{72 \times 10^{-3}}{9.8 \times 10^{-1}}\end$

$r^2$ $\begin{equation}=7.34 \times 10^{-2} \mathrm{metre}\end$

$r=2.70924 \times 10^{-1} metre.$

Thus, the charge B should be $2.70924 \times 10^{-1} metre$  away from charge A.