Question

Two metal spheres of identical mass m 3.60 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a charge of 0.805 μC, and the right-hand sphere carries a charge of 1.47 μc. what is the equilibrium separation between the centers of the two spheres? 1.030 Your response differs from the correct answer by more than 10%. Double check your calculations. m

Answer 1

Answer:0.67 m

Step-by-step explanation:

Given

mass of sphere(m)=3.60 gm

Length of String =0.5 m

Left hand charge (
`q_1`)
`=0.805 \mu c`

Right hand charge (
`q_2`)
`=1.47 \mu c`

From Free body diagram

`Tcos\theta =mg------1`

`Tsin\theta =F-------2`

Divide 2 & 1 we get

`tan\theta =(F)/(mg)`

Where F is given Electrostatic force between two charge

`F=(kq_1q_2)/(d^2)`

d=distance between them

`d=2Lsin\theta`

Here
`\theta` is very small

therefore

`tan\theta \approx sin\theta =(d)/(2L)`

thus

`d^3=(kq_1q_22L)/(mg)`

`d^3=(9* 10^9* 0.805* 1.47* 10^(-12)* 2* 0.5)/(3.6* 10^(-3)* 9.81)`

d=0.67 m