Question

G a small metal sphere, carrying a net charge is held stationary. what is the speed are 0.4 m apart

Answer 1

Answer:

The final speed of small metal sphere is 12.6 m/s.

Explanation:

Given that,

Distance = 0.4 m

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the final speed of small metal sphere

Using conservation of energy

$\dfrac{1}{2}mv_{1}^2+\dfrac{kq_{1}q_{2}}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{kq_{1}q_{2}}{r_{2}}$

$\dfrac{1}{2}m(v_{1}^2-v_{2}^2)=kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{2}^2)=9\times10^{9}\times(-2)\times10^{-6}\times(-8)\times10^{-6}(\dfrac{1}{0.4}-\dfrac{1}{0.8})$

$400-v_{2}^2=240$

$-v_{2}^2=240-400$

$v_{2}=\sqrt{160}\ m/s$

$v_{2}=12.6\ m/s$

Hence, The final speed of small metal sphere is 12.6 m/s.