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

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.

Step-by-step 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=(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

`(kQ_1q)/(x^2)-(kQ_2q)/((d-x)^2)=0\\(k* 60 q)/(x^2)-(k6.66q)/((2.5-x)^2)=0\\\\x=0.75* d\\\\x=0.75* 2.5\\\\x=1.88\ \rm cm`.

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