Question

A Particle A having charge q and mass m' is placed at the bottom of smooth inclined plane inclination theta Where should a block B, having same Charge and mass, be placed on the incline, so that it may remain in equilibrium ?​

Answer 1

The distance on the incline, where the other block should be placed to bein equilibrium is

$\boxed{q\sqrt{\frac{k}{mg\sin\theta}}}$

Explanation:

In the case of equilibrium, the Coulomb's force will be equal to the component of the weight of the particle of mass m along the inclined plane

We know that

Force acting a charge $q_1$ by another charge $q_2$, separated by a distance $r$ is given by

$F=k\frac{q_1q_2}{r^2}$        (where k is coulomb's constant)

Here, both the charges are $q$ and we need to find $r$

The component of the weight $mg$ along the inclined plane will be

$mg\sin\theta$

Thus,

$mg\sin\theta=k\frac{q^2}{r^2}$

$\implies r^2=\frac{kq^2}{mg\sin\theta}$

$\implies r=\sqrt{\frac{kq^2}{mg\sin\theta}}$

$\implies r=q\sqrt{\frac{k}{mg\sin\theta}}$

Hope this answer is helpful.

Know More:

Q: 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?

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