Question

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Answer 1

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

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 speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

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

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

Put the value into the formula

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

$0.00075(400-v_{f}^2)=0.18$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

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

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$