# HCV2 Pg 122 Q30

## HCV2 Pg 122 Q30

A particle A having a charge of 2*10^-6 C and a mass of 100g is placed at the bottom of a smooth inclined plane of inclination 30. Where should another particle B having same charge and mass be placed on the incline so that it may remain in equilibrium?

P.S: Well my main confusion in this sum is that why are we equating mgsin30 with F and not mgcos30?

P.S: Well my main confusion in this sum is that why are we equating mgsin30 with F and not mgcos30?

**Shohini**- Posts : 15

Join date : 2008-05-31

Age : 28

## Re: HCV2 Pg 122 Q30

Well the problem reads simple. It says that one charge particle is at the bottom of the plane, which is inclined to the horizontal at an angle of 30 degrees. Since the gravitational force on the charge particle on the plane is going to mgsin 30, which has to balance the electrostatic force (repulsive). Therefore the problem is of now the equation

k_e*q^2/r^2 = mgsin 30

the component of force (gravitational) along the plane is mgsin 30 and not mgcos 30

where r is the distance along the plane

plugging the given values the question is done

hope this helps!

Pinnaka

k_e*q^2/r^2 = mgsin 30

the component of force (gravitational) along the plane is mgsin 30 and not mgcos 30

where r is the distance along the plane

plugging the given values the question is done

hope this helps!

Pinnaka

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

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