Question

Two small spherical insulators separated by 2.5 cm, which is much greater than either of their diameters. Both carry positive charge, one +60.0 microCoulombs and the other +6.66 microCoulombs. A third positive charge remains at rest between the two spheres and along the line joining them. What is the position of this charged sphere?

Answer 1

Answer:

The third charge is placed at a distance of 1.88 cm  from the centre of 60 micro coulombs charge to the right of it.

Explanation:

Given:

Distance between the centre of insulators=d=2.5 cm

Charge on insulator(1`)= 60 $\rm \mu C$

Charge on insulator(2`)= 6.66 $\rm \mu C$

Let third charge be placed at a distance of x xm from the centre of 60 $\rm \mu C$ charge.

Now according to coulombs law the force between tow charges placed at distance r is given by

$F=\dfrac{kQ_1Q_2}{r^2}$

• where $Q_1, Q_2$ are the magnitude of the charges.
• r is the radial distance between their centres.

Let magnitude of third charge be q and it placed at a distance of x from  higher magnitude charge. According to question the net force on third charge is zero so we have

$\dfrac{kQ_1q}{x^2}-\dfrac{kQ_2q}{(d-x)^2}=0\\\dfrac{k\times 60 q}{x^2}-\dfrac{k6.66q}{(2.5-x)^2}=0\\\\x=0.75\times d\\\\x=0.75\times 2.5\\\\x=1.88\ \rm cm$.

Hence the position of third charge is calculated in order to be rest.