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

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

$tan\theta =\frac{F}{mg}$

Where F is given Electrostatic force between two charge

$F=\frac{kq_1q_2}{d^2}$

d=distance between them

$d=2Lsin\theta$

Here $\theta$ is very small

therefore

$tan\theta \approx sin\theta =\frac{d}{2L}$

thus

$d^3=\frac{kq_1q_22L}{mg}$

$d^3=\frac{9\times 10^9\times 0.805\times 1.47\times 10^{-12}\times 2\times 0.5}{3.6\times 10^{-3}\times 9.81}$

d=0.67 m